Php как получить ошибку mysql

(PHP 4, PHP 5)

mysql_errorВозвращает текст ошибки последней операции с MySQL

Описание

mysql_error(resource $link_identifier = NULL): string

Список параметров

link_identifier

Соединение MySQL. Если идентификатор соединения не был указан,
используется последнее соединение, открытое mysql_connect(). Если такое соединение не было найдено,
функция попытается создать таковое, как если бы mysql_connect() была вызвана без параметров.
Если соединение не было найдено и не смогло быть создано, генерируется ошибка уровня E_WARNING.

Возвращаемые значения

Возвращает текст ошибки выполнения последней функции MySQL,
или '' (пустую строку), если операция
выполнена успешно.

Примеры

Пример #1 Пример использования mysql_error()


<?php
$link
= mysql_connect("localhost", "mysql_user", "mysql_password");mysql_select_db("nonexistentdb", $link);
echo
mysql_errno($link) . ": " . mysql_error($link). "n";mysql_select_db("kossu", $link);
mysql_query("SELECT * FROM nonexistenttable", $link);
echo
mysql_errno($link) . ": " . mysql_error($link) . "n";
?>

Результатом выполнения данного примера
будет что-то подобное:

1049: Unknown database 'nonexistentdb'
1146: Table 'kossu.nonexistenttable' doesn't exist

aleczapka _at) gmx dot net

18 years ago


If you want to display errors like "Access denied...", when mysql_error() returns "" and mysql_errno() returns 0, use  $php_errormsg. This Warning will be stored there.  You need to have track_errors set to true in your php.ini.

Note. There is a bug in either documentation about error_reporting() or in mysql_error() function cause manual for mysql_error(), says:  "Errors coming back from the MySQL database backend no longer issue warnings." Which is not true.


Florian Sidler

13 years ago


Be aware that if you are using multiple MySQL connections you MUST support the link identifier to the mysql_error() function. Otherwise your error message will be blank.

Just spent a good 30 minutes trying to figure out why i didn't see my SQL errors.


Pendragon Castle

14 years ago


Using a manipulation of josh ><>'s function, I created the following. It's purpose is to use the DB to store errors. It handles both original query, as well as the error log. Included Larry Ullman's escape_data() as well since I use it in q().

<?php

function escape_data($data){

global
$dbc;

if(
ini_get('magic_quotes_gpc')){

 
$data=stripslashes($data);

}

return
mysql_real_escape_string(trim($data),$dbc);

}

function

q($page,$query){

// $page

$result = mysql_query($query);

if (
mysql_errno()) {

 
$error = "MySQL error ".mysql_errno().": ".mysql_error()."n<br>When executing:<br>n$queryn<br>";

 
$log = mysql_query("INSERT INTO db_errors (error_page,error_text) VALUES ('$page','".escape_data($error)."')");

}

}
// Run the query using q()

$query = "INSERT INTO names (first, last) VALUES ('myfirst', 'mylast'");

$result = q("Sample Page Title",$query);

?>


l dot poot at twing dot nl

16 years ago


When creating large applications it's quite handy to create a custom function for handling queries. Just include this function in every script. And use db_query(in this example) instead of mysql_query.

This example prompts an error in debugmode (variable $b_debugmode ). An e-mail with the error will be sent to the site operator otherwise.

The script writes a log file in directory ( in this case /log ) as well.

The system is vulnerable when database/query information is prompted to visitors. So be sure to hide this information for visitors anytime.

Regars,

Lennart Poot
http://www.twing.nl

<?php
$b_debugmode
= 1; // 0 || 1$system_operator_mail = 'developer@company.com';
$system_from_mail = 'info@mywebsite.com';

function

db_query( $query ){
  global
$b_debugmode;// Perform Query
 
$result = mysql_query($query);// Check result
  // This shows the actual query sent to MySQL, and the error. Useful for debugging.
 
if (!$result) {
    if(
$b_debugmode){
     
$message  = '<b>Invalid query:</b><br>' . mysql_error() . '<br><br>';
     
$message .= '<b>Whole query:</b><br>' . $query . '<br><br>';
      die(
$message);
    }
raise_error('db_query_error: ' . $message);
  }
  return
$result;
}

  function

raise_error( $message ){
    global
$system_operator_mail, $system_from_mail;$serror=
   
"Env:       " . $_SERVER['SERVER_NAME'] . "rn" .
   
"timestamp: " . Date('m/d/Y H:i:s') . "rn" .
   
"script:    " . $_SERVER['PHP_SELF'] . "rn" .
   
"error:     " . $message ."rnrn";// open a log file and write error
   
$fhandle = fopen( '/logs/errors'.date('Ymd').'.txt', 'a' );
    if(
$fhandle){
     
fwrite( $fhandle, $serror );
     
fclose(( $fhandle ));
     }
// e-mail error to system operator
   
if(!$b_debugmode)
     
mail($system_operator_mail, 'error: '.$message, $serror, 'From: ' . $system_from_mail );
  }
?>


Anonymous

18 years ago


My suggested implementation of mysql_error():

$result = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.n<br />Query: " . $query . "<br />nError: (" . mysql_errno() . ") " . mysql_error());

This will print out something like...

A fatal MySQL error occured.

Query: SELECT * FROM table

Error: (err_no) Bla bla bla, you did everything wrong

It's very useful to see your query in order to detect problems with syntax. Most often, the output message from MySQL doesn't let you see enough of the query in the error message to let you see where your query went bad- it a missing quote, comma, or ( or ) could have occured well before the error was detected. I do -not- recomend using this procedure, however, for queries which execute on your site that are not user-specific as it has the potential to leak sensative data. Recomended use is just for debugging/building a script, and for general user-specific queries which would at the worst, leak the users own information to themself.

Good luck,

-Scott


olaf at amen-online dot de

18 years ago


When dealing with user input, make sure that you use
<?php
echo htmlspecialchars (mysql_error ());
?>
instead of
<?php
echo mysql_error ();
?>

Otherwise it might be possible to crack into your system by submitting data that causes the SQL query to fail and that also contains javascript commands.

Would it make sense to change the examples in the documentation for mysql_query () and for mysql_error () accordingly?


Anonymous

22 years ago


some error can't handle. Example:

ERROR 1044: Access denied for user: 'ituser@mail.ramon.intranet' to database 'itcom'

This error ocurrs when a intent of a sql insert of no authorized user. The results: mysql_errno = 0 and the mysql_error = "" .


Gianluigi_Zanettini-MegaLab.it

16 years ago


"Errors coming back from the MySQL database backend no longer issue warnings." Please note, you have an error/bug here. In fact, MySQL 5.1 with PHP 5.2:

Warning: mysql_connect() [function.mysql-connect]: Unknown MySQL server host 'locallllllhost' (11001)

That's a warning, which is not trapped by mysql_error()!


scott at rocketpack dot net

19 years ago


My suggested implementation of mysql_error():

$result = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.n<br />Query: " . $query . "<br />nError: (" . mysql_errno() . ") " . mysql_error());

This will print out something like...

<b>A fatal MySQL error occured</b>.
Query: SELECT * FROM table
Error: (err_no) Bla bla bla, you did everything wrong

It's very useful to see your query in order to detect problems with syntax. Most often, the output message from MySQL doesn't let you see enough of the query in the error message to let you see where your query went bad- it a missing quote, comma, or ( or ) could have occured well before the error was detected. I do -not- recomend using this procedure, however, for queries which execute on your site that are not user-specific as it has the potential to leak sensative data. Recomended use is just for debugging/building a script, and for general user-specific queries which would at the worst, leak the users own information to themself.

Good luck,

-Scott


josh ><>

19 years ago


Oops, the code in my previous post only works for queries that don't return data (INSERT, UPDATE, DELETE, etc.), this updated function should work for all types of queries (using $result = myquery($query);):

    function myquery ($query) {
        $result = mysql_query($query);
        if (mysql_errno())
            echo "MySQL error ".mysql_errno().": ".mysql_error()."n<br>When executing:<br>n$queryn<br>";
        return $result;
    }


phpnet at robzazueta dot com

16 years ago


This is a big one - As of MySQL 4.1 and above, apparently, the way passwords are hashed has changed. PHP 4.x is not compatible with this change, though PHP 5.0 is. I'm still using the 4.x series for various compatibility reasons, so when I set up MySQL 5.0.x on IIS 6.0 running PHP 4.4.4 I was surpised to get this error from mysql_error():

MYSQL: Client does not support authentication protocol requested by server; consider upgrading MySQL client

According to the MySQL site (http://dev.mysql.com/doc/refman/5.0/en/old-client.html) the best fix for this is to use the OLD_PASSWORD() function for your mysql DB user. You can reset it by issuing to MySQL:

Set PASSWORD for 'user'@'host' = OLD_PASSWORD('password');

This saved my hide.


miko_il AT yahoo DOT com

19 years ago

Gianluigi_Zanettini-MegaLab.it

16 years ago


A friend of mine proposed a great solution.

<?php
$old_track
= ini_set('track_errors', '1');

.....

        if (

$this->db_handle!=FALSE && $db_selection_status!=FALSE)
            {
           
$this->connected=1;
           
ini_set('track_errors', $old_track);
            }
        else
            {
           
$this->connected=-1;
           
$mysql_warning=$php_errormsg;
           
ini_set('track_errors', $old_track);
            throw new
mysql_cns_exception(1, $mysql_warning . " " . mysql_error());
            }
?>


Gerrit

8 years ago


The following code returns two times the same error, even though I would have expected only one:

   $ conn = mysql_connect ('localhost', 'root', '');
   $ conn2 = mysql_connect ('localhost', 'root', '');

     mysql_select_db ('db1', $ conn);
   mysql_select_db ('db2', $ conn2);

     $ result = mysql_query ("select 1 from dual", $ conn);
   $ result2 = mysql_query ("select 1 from luad", $ conn2);

     echo mysql_error ($ conn) "<hr>".
   echo mysql_error ($ conn2) "<hr>".

The reason for this is that mysql_connect not working as expected a further connection returns. Since the parameters are equal, a further reference to the previous link is returned. So also changes the second mysql_select_db the selected DB of $conn to 'db2'.

If you change the connection parameters of the second connection to 127.0.0.1, a new connection is returned. In addition to the parameters new_link the mysql_connect() function to be forced.


mysql_error

(PHP 4, PHP 5)

mysql_errorВозвращает текст ошибки последней операции с MySQL

Описание

string mysql_error
([ resource $link_identifier = NULL
] )

Возвращает текст ошибки выполнения последней функции MySQL.
Ошибки работы с MySQL больше не вызывают сообщений в PHP. Вместо
этого используйте функцию mysql_error(), для
получения сообщения об ошибке. Учтите, что функция возвращает текст
ошибки только последней выполненной функции MySQL (исключая
mysql_error() и mysql_errno()),
поэтому убедитесь, что вы вызываете данную функцию до вызова
следующей функции MySQL.

Список параметров


link_identifier

Соединение MySQL. Если идентификатор соединения не был указан,
используется последнее соединение, открытое mysql_connect(). Если такое соединение не было найдено,
функция попытается создать таковое, как если бы mysql_connect() была вызвана без параметров.
Если соединение не было найдено и не смогло быть создано, генерируется ошибка уровня E_WARNING.

Возвращаемые значения

Возвращает текст ошибки выполнения последней функции MySQL,
или » (пустую строку), если операция
выполнена успешно.

Примеры

Пример #1 Пример использования mysql_error()


<?php
$link 
mysql_connect("localhost""mysql_user""mysql_password");mysql_select_db("nonexistentdb"$link);
echo 
mysql_errno($link) . ": " mysql_error($link). "n";mysql_select_db("kossu"$link);
mysql_query("SELECT * FROM nonexistenttable"$link);
echo 
mysql_errno($link) . ": " mysql_error($link) . "n";
?>

Результатом выполнения данного примера
будет что-то подобное:

1049: Unknown database 'nonexistentdb'
1146: Table 'kossu.nonexistenttable' doesn't exist

Вернуться к: MySQL

I have a mysqli query that wont execute and I would like to display information about why that’s happening. Im just fooling around with this example but I imagine something like this:

$myQuery= $mysqli->query("UPDATE table SET id = 1 WHERE id = 3");
if(!$myQuery) //If query couldnt be executed
{
echo $mysqli->error; //Display information about why wasnt executed (eg. Error: couldnt find table)
}

asked Apr 10, 2017 at 4:49

nightlynutria's user avatar

1

try using

// Perform a query, check for error
if (!mysqli_query($con,"UPDATE table SET id = 1 WHERE id = 3"))
{
 echo("Error description: " . mysqli_error($con));
}

mysqli_close($con);

answered Apr 10, 2017 at 4:53

Nishant Nair's user avatar

Nishant NairNishant Nair

1,9891 gold badge13 silver badges18 bronze badges

0

PHP | mysqli_error() Function

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    The mysqli_error() function is used to return the error in the most recent MySQL function call that failed. If there are multiple MySQL function calls, the error in the last statement is the one that is pointed out by the function.

    Syntax:

    mysqli_error("database_name")
    

    Parameters: This function accepts single parameter as mentioned above and described below:

    • database_name: It is the database on which operations are being performed. It is a mandatory parameter.

    Program 1:

    <?php

    $conn = mysqli_connect(

        "localhost", "root", "", "Persons"); 

    if (mysqli_connect_errno()) { 

        echo "Database connection failed."

    if (!mysqli_query($link, "SET Age=1")) {

        printf("Error message: %sn", mysqli_error($conn));

    }    

    mysqli_close($conn);

    ?>

    Suppose the operation is being carried out on the table given below:

    The output will be:

    Error message: Unknown system variable 'Age'
    

    Program 2:

    <?php

    $conn = mysqli_connect(

        "localhost", "root", "", "Persons"); 

    if (mysqli_connect_errno()) { 

        echo "Database connection failed."

    if (!mysqli_query($link, "SET Firstname='Arkadyuti'")) {

        printf("Error message: %sn", mysqli_error());

    }    

    mysqli_close($conn);

    ?>

    Output:

    Error message: mysqli_error() expects exactly 1 parameter, 0 given
    

    This example also demonstrates that mysqli_error() needs a database as a parameter.

    Last Updated :
    23 Apr, 2020

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    (PHP 4, PHP 5)

    mysql_errorВозвращает текст ошибки последней операции с MySQL

    Описание

    mysql_error(resource $link_identifier = NULL): string

    Список параметров

    link_identifier

    Соединение MySQL. Если идентификатор соединения не был указан,
    используется последнее соединение, открытое mysql_connect(). Если такое соединение не было найдено,
    функция попытается создать таковое, как если бы mysql_connect() была вызвана без параметров.
    Если соединение не было найдено и не смогло быть создано, генерируется ошибка уровня E_WARNING.

    Возвращаемые значения

    Возвращает текст ошибки выполнения последней функции MySQL,
    или '' (пустую строку), если операция
    выполнена успешно.

    Примеры

    Пример #1 Пример использования mysql_error()


    <?php
    $link
    = mysql_connect("localhost", "mysql_user", "mysql_password");mysql_select_db("nonexistentdb", $link);
    echo
    mysql_errno($link) . ": " . mysql_error($link). "n";mysql_select_db("kossu", $link);
    mysql_query("SELECT * FROM nonexistenttable", $link);
    echo
    mysql_errno($link) . ": " . mysql_error($link) . "n";
    ?>

    Результатом выполнения данного примера
    будет что-то подобное:

    1049: Unknown database 'nonexistentdb'
    1146: Table 'kossu.nonexistenttable' doesn't exist
    

    aleczapka _at) gmx dot net

    18 years ago


    If you want to display errors like "Access denied...", when mysql_error() returns "" and mysql_errno() returns 0, use  $php_errormsg. This Warning will be stored there.  You need to have track_errors set to true in your php.ini.

    Note. There is a bug in either documentation about error_reporting() or in mysql_error() function cause manual for mysql_error(), says:  "Errors coming back from the MySQL database backend no longer issue warnings." Which is not true.

    Florian Sidler

    12 years ago


    Be aware that if you are using multiple MySQL connections you MUST support the link identifier to the mysql_error() function. Otherwise your error message will be blank.

    Just spent a good 30 minutes trying to figure out why i didn't see my SQL errors.

    Pendragon Castle

    14 years ago


    Using a manipulation of josh ><>'s function, I created the following. It's purpose is to use the DB to store errors. It handles both original query, as well as the error log. Included Larry Ullman's escape_data() as well since I use it in q().

    <?phpfunction escape_data($data){

    global

    $dbc;

    if(

    ini_get('magic_quotes_gpc')){$data=stripslashes($data);

    }

    return

    mysql_real_escape_string(trim($data),$dbc);

    }

    function

    q($page,$query){// $page$result = mysql_query($query);

    if (

    mysql_errno()) {$error = "MySQL error ".mysql_errno().": ".mysql_error()."n<br>When executing:<br>n$queryn<br>";$log = mysql_query("INSERT INTO db_errors (error_page,error_text) VALUES ('$page','".escape_data($error)."')");

    }

    }

    // Run the query using q()$query = "INSERT INTO names (first, last) VALUES ('myfirst', 'mylast'");$result = q("Sample Page Title",$query);?>

    l dot poot at twing dot nl

    16 years ago


    When creating large applications it's quite handy to create a custom function for handling queries. Just include this function in every script. And use db_query(in this example) instead of mysql_query.

    This example prompts an error in debugmode (variable $b_debugmode ). An e-mail with the error will be sent to the site operator otherwise.

    The script writes a log file in directory ( in this case /log ) as well.

    The system is vulnerable when database/query information is prompted to visitors. So be sure to hide this information for visitors anytime.

    Regars,

    Lennart Poot
    http://www.twing.nl

    <?php
    $b_debugmode
    = 1; // 0 || 1$system_operator_mail = 'developer@company.com';
    $system_from_mail = 'info@mywebsite.com';

    function

    db_query( $query ){
      global
    $b_debugmode;// Perform Query
     
    $result = mysql_query($query);// Check result
      // This shows the actual query sent to MySQL, and the error. Useful for debugging.
     
    if (!$result) {
        if(
    $b_debugmode){
         
    $message  = '<b>Invalid query:</b><br>' . mysql_error() . '<br><br>';
         
    $message .= '<b>Whole query:</b><br>' . $query . '<br><br>';
          die(
    $message);
        }
    raise_error('db_query_error: ' . $message);
      }
      return
    $result;
    }

      function

    raise_error( $message ){
        global
    $system_operator_mail, $system_from_mail;$serror=
       
    "Env:       " . $_SERVER['SERVER_NAME'] . "rn" .
       
    "timestamp: " . Date('m/d/Y H:i:s') . "rn" .
       
    "script:    " . $_SERVER['PHP_SELF'] . "rn" .
       
    "error:     " . $message ."rnrn";// open a log file and write error
       
    $fhandle = fopen( '/logs/errors'.date('Ymd').'.txt', 'a' );
        if(
    $fhandle){
         
    fwrite( $fhandle, $serror );
         
    fclose(( $fhandle ));
         }
    // e-mail error to system operator
       
    if(!$b_debugmode)
         
    mail($system_operator_mail, 'error: '.$message, $serror, 'From: ' . $system_from_mail );
      }
    ?>

    Anonymous

    18 years ago


    My suggested implementation of mysql_error():

    $result = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.n<br />Query: " . $query . "<br />nError: (" . mysql_errno() . ") " . mysql_error());

    This will print out something like...

    A fatal MySQL error occured.

    Query: SELECT * FROM table

    Error: (err_no) Bla bla bla, you did everything wrong

    It's very useful to see your query in order to detect problems with syntax. Most often, the output message from MySQL doesn't let you see enough of the query in the error message to let you see where your query went bad- it a missing quote, comma, or ( or ) could have occured well before the error was detected. I do -not- recomend using this procedure, however, for queries which execute on your site that are not user-specific as it has the potential to leak sensative data. Recomended use is just for debugging/building a script, and for general user-specific queries which would at the worst, leak the users own information to themself.

    Good luck,

    -Scott

    olaf at amen-online dot de

    18 years ago


    When dealing with user input, make sure that you use
    <?php
    echo htmlspecialchars (mysql_error ());
    ?>
    instead of
    <?php
    echo mysql_error ();
    ?>

    Otherwise it might be possible to crack into your system by submitting data that causes the SQL query to fail and that also contains javascript commands.

    Would it make sense to change the examples in the documentation for mysql_query () and for mysql_error () accordingly?

    Anonymous

    21 years ago


    some error can't handle. Example:

    ERROR 1044: Access denied for user: 'ituser@mail.ramon.intranet' to database 'itcom'

    This error ocurrs when a intent of a sql insert of no authorized user. The results: mysql_errno = 0 and the mysql_error = "" .

    Gianluigi_Zanettini-MegaLab.it

    15 years ago


    "Errors coming back from the MySQL database backend no longer issue warnings." Please note, you have an error/bug here. In fact, MySQL 5.1 with PHP 5.2:

    Warning: mysql_connect() [function.mysql-connect]: Unknown MySQL server host 'locallllllhost' (11001)

    That's a warning, which is not trapped by mysql_error()!

    scott at rocketpack dot net

    19 years ago


    My suggested implementation of mysql_error():

    $result = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.n<br />Query: " . $query . "<br />nError: (" . mysql_errno() . ") " . mysql_error());

    This will print out something like...

    <b>A fatal MySQL error occured</b>.
    Query: SELECT * FROM table
    Error: (err_no) Bla bla bla, you did everything wrong

    It's very useful to see your query in order to detect problems with syntax. Most often, the output message from MySQL doesn't let you see enough of the query in the error message to let you see where your query went bad- it a missing quote, comma, or ( or ) could have occured well before the error was detected. I do -not- recomend using this procedure, however, for queries which execute on your site that are not user-specific as it has the potential to leak sensative data. Recomended use is just for debugging/building a script, and for general user-specific queries which would at the worst, leak the users own information to themself.

    Good luck,

    -Scott

    josh ><>

    19 years ago


    Oops, the code in my previous post only works for queries that don't return data (INSERT, UPDATE, DELETE, etc.), this updated function should work for all types of queries (using $result = myquery($query);):

        function myquery ($query) {
            $result = mysql_query($query);
            if (mysql_errno())
                echo "MySQL error ".mysql_errno().": ".mysql_error()."n<br>When executing:<br>n$queryn<br>";
            return $result;
        }

    phpnet at robzazueta dot com

    16 years ago


    This is a big one - As of MySQL 4.1 and above, apparently, the way passwords are hashed has changed. PHP 4.x is not compatible with this change, though PHP 5.0 is. I'm still using the 4.x series for various compatibility reasons, so when I set up MySQL 5.0.x on IIS 6.0 running PHP 4.4.4 I was surpised to get this error from mysql_error():

    MYSQL: Client does not support authentication protocol requested by server; consider upgrading MySQL client

    According to the MySQL site (http://dev.mysql.com/doc/refman/5.0/en/old-client.html) the best fix for this is to use the OLD_PASSWORD() function for your mysql DB user. You can reset it by issuing to MySQL:

    Set PASSWORD for 'user'@'host' = OLD_PASSWORD('password');

    This saved my hide.

    miko_il AT yahoo DOT com

    19 years ago

    Gianluigi_Zanettini-MegaLab.it

    15 years ago


    A friend of mine proposed a great solution.

    <?php
    $old_track
    = ini_set('track_errors', '1');

    .....

            if (

    $this->db_handle!=FALSE && $db_selection_status!=FALSE)
                {
               
    $this->connected=1;
               
    ini_set('track_errors', $old_track);
                }
            else
                {
               
    $this->connected=-1;
               
    $mysql_warning=$php_errormsg;
               
    ini_set('track_errors', $old_track);
                throw new
    mysql_cns_exception(1, $mysql_warning . " " . mysql_error());
                }
    ?>

    Gerrit

    8 years ago


    The following code returns two times the same error, even though I would have expected only one:

       $ conn = mysql_connect ('localhost', 'root', '');
       $ conn2 = mysql_connect ('localhost', 'root', '');

         mysql_select_db ('db1', $ conn);
       mysql_select_db ('db2', $ conn2);

         $ result = mysql_query ("select 1 from dual", $ conn);
       $ result2 = mysql_query ("select 1 from luad", $ conn2);

         echo mysql_error ($ conn) "<hr>".
       echo mysql_error ($ conn2) "<hr>".

    The reason for this is that mysql_connect not working as expected a further connection returns. Since the parameters are equal, a further reference to the previous link is returned. So also changes the second mysql_select_db the selected DB of $conn to 'db2'.

    If you change the connection parameters of the second connection to 127.0.0.1, a new connection is returned. In addition to the parameters new_link the mysql_connect() function to be forced.

    mysqli_error

    (PHP 5, PHP 7, PHP 8)

    mysqli::$errormysqli_errorВозвращает строку с описанием последней ошибки

    Описание

    Объектно-ориентированный стиль

    Процедурный стиль

    mysqli_error(mysqli $mysql): string

    Возвращаемые значения

    Строка с описанием ошибки. Пустая строка, если ошибки нет.

    Примеры

    Пример #1 Пример с $mysqli->error

    Объектно-ориентированный стиль


    <?php
    $mysqli
    = new mysqli("localhost", "my_user", "my_password", "world");/* Проверить соединение */
    if ($mysqli->connect_errno) {
    printf("Соединение не удалось: %sn", $mysqli->connect_error);
    exit();
    }

    if (!

    $mysqli->query("SET a=1")) {
    printf("Сообщение ошибки: %sn", $mysqli->error);
    }
    /* Закрыть соединение */
    $mysqli->close();
    ?>

    Процедурный стиль


    <?php
    $link
    = mysqli_connect("localhost", "my_user", "my_password", "world");/* Проверить соединение */
    if (mysqli_connect_errno()) {
    printf("Соединение не удалось: %sn", mysqli_connect_error());
    exit();
    }

    if (!

    mysqli_query($link, "SET a=1")) {
    printf("Сообщение ошибки: %sn", mysqli_error($link));
    }
    /* Закрыть соединение */
    mysqli_close($link);
    ?>

    Результат выполнения данных примеров:

    Сообщение ошибки: Unknown system variable 'a'
    

    Смотрите также

    • mysqli_connect_errno() — Возвращает код ошибки последней попытки соединения
    • mysqli_connect_error() — Возвращает описание последней ошибки подключения
    • mysqli_errno() — Возвращает код ошибки последнего вызова функции
    • mysqli_sqlstate() — Возвращает код состояния SQLSTATE последней MySQL операции

    information at saunderswebsolutions dot com

    17 years ago


    The mysqli_sql_exception class is not available to PHP 5.05

    I used this code to catch errors
    <?php
    $query
    = "SELECT XXname FROM customer_table ";
    $res = $mysqli->query($query);

    if (!

    $res) {
      
    printf("Errormessage: %sn", $mysqli->error);
    }
    ?>
    The problem with this is that valid values for $res are: a mysqli_result object , true or false
    This doesn't tell us that there has been an error with the sql used.
    If you pass an update statement, false is a valid result if the update fails.

    So, a better way is:
    <?php
    $query
    = "SELECT XXname FROM customer_table ";
    $res = $mysqli->query($query);

    if (!

    $mysqli->error) {
      
    printf("Errormessage: %sn", $mysqli->error);
    }
    ?>

    This would output something like:
    Unexpected PHP error [mysqli::query() [<a href='function.query'>function.query</a>]: (42S22/1054): Unknown column 'XXname' in 'field list'] severity [E_WARNING] in [G:database.php] line [249]

    Very frustrating as I wanted to also catch the sql error and print out the stack trace.

    A better way is:

    <?php
    mysqli_report
    (MYSQLI_REPORT_OFF); //Turn off irritating default messages$mysqli = new mysqli("localhost", "my_user", "my_password", "world");$query = "SELECT XXname FROM customer_table ";
    $res = $mysqli->query($query);

    if (

    $mysqli->error) {
        try {   
            throw new
    Exception("MySQL error $mysqli->error <br> Query:<br> $query", $msqli->errno);   
        } catch(
    Exception $e ) {
            echo
    "Error No: ".$e->getCode(). " - ". $e->getMessage() . "<br >";
            echo
    nl2br($e->getTraceAsString());
        }
    }
    //Do stuff with the result
    ?>
    Prints out something like:
    Error No: 1054
    Unknown column 'XXname' in 'field list'
    Query:
    SELECT XXname FROM customer_table

    #0 G:database.php(251): database->dbError('Unknown column ...', 1054, 'getQuery()', 'SELECT XXname F...')
    #1 G:dataWorkSites1framework5testsdbtest.php(29): database->getString('SELECT XXname F...')
    #2 c:PHPincludessimpletestrunner.php(58): testOfDB->testGetVal()
    #3 c:PHPincludessimpletestrunner.php(96): SimpleInvoker->invoke('testGetVal')
    #4 c:PHPincludessimpletestrunner.php(125): SimpleInvokerDecorator->invoke('testGetVal')
    #5 c:PHPincludessimpletestrunner.php(183): SimpleErrorTrappingInvoker->invoke('testGetVal')
    #6 c:PHPincludessimpletestsimple_test.php(90): SimpleRunner->run()
    #7 c:PHPincludessimpletestsimple_test.php(498): SimpleTestCase->run(Object(HtmlReporter))
    #8 c:PHPincludessimpletestsimple_test.php(500): GroupTest->run(Object(HtmlReporter))
    #9 G:all_tests.php(16): GroupTest->run(Object(HtmlReporter))

    This will actually print out the error, a stack trace and the offending sql statement. Much more helpful when the sql statement is generated somewhere else in the code.

    se (at) brainbits (dot) net

    16 years ago


    The decription "mysqli_error -- Returns a string description of the LAST error" is not exactly that what you get from mysqli_error. You get the error description from the last mysqli-function, not from the last mysql-error.

    If you have the following situation

    if (!$mysqli->query("SET a=1")) {
       $mysqli->query("ROLLBACK;")
       printf("Errormessage: %sn", $mysqli->error);
    }

    you don't get an error-message, if the ROLLBACK-Query didn't failed, too. In order to get the right error-message you have to write:

    if (!$mysqli->query("SET a=1")) {
       printf("Errormessage: %sn", $mysqli->error);
       $mysqli->query("ROLLBACK;")
    }

    callforeach at gmail dot com

    7 years ago


    I had to set mysqli_report(MYSQLI_REPORT_ALL) at the begin of my script to be able to catch mysqli errors within the catch block of my php code.

    Initially, I used the below code to throw and subsequent catch mysqli exceptions

    <?php
    try {
      
    $mysqli = new mysqli('localhost','root','pwd','db');
        if (
    $mysqli->connect_errno)
            throw new
    Exception($mysqli->connect_error);

    } catch (

    Exception $e) {
         echo
    $e->getMessage();
    }
    I realized the exception was being thrown before the actual throw statement and hence the catch block was not being called.My current code looks like
    mysqli_report
    (MYSQLI_REPORT_ALL) ;
    try {
         
    $mysqli = new mysqli('localhost','root','pwd','db');
         
    /* I don't need to throw the exception, it's being thrown automatically */} catch (Exception $e) {
      echo
    $e->getMessage();
    }
    This works fine and I'm able to trap all mysqli errors

    abderrahmanekaddour dot aissat at gmail dot com

    5 months ago


    <?php// The idea is the add formated errors information for developers to easier bugs detection.$myfile = fopen("database_log.log", "r");
    $db = new mysqli("localhost", "root","root","data");
    if(!
    $db->query("SELECT")){
     
    $timestamp = new DateTime();
     
    $data_err = " {
         "title": " Select statement error ",
         "date_time": "
    .$timestamp->getTimestamp().",
         "error":" "
    .$db->error." "
         } "
    ; // Do more information
     
    fwrite($myfile, $data_err); // writing data
    }
       
    // In separate file do file read and format it for good visual.$db->close(); 
    fclose($myfile);
    ?>

    asmith16 at littlesvr dot ca

    9 years ago


    Please note that the string returned may contain data initially provided by the user, possibly making your code vulnerable to XSS.

    So even if you escape everything in your SQL query using mysqli_real_escape_string(), make sure that if you plan to display the string returned by mysqli_error() you run that string through htmlspecialchars().

    As far as I can tell the two escape functions don't escape the same characters, which is why you need both (the first for SQL and the second for HTML/JS).

    information at saunderswebsolutions dot com

    17 years ago


    Hi, you can also use the new mysqli_sql_exception to catch sql errors.
    Example:
    <?php
    //set up $mysqli_instance here..
    $Select = "SELECT xyz FROM mytable ";
    try {
       
    $res = $mysqli_instance->query($Select);
    }catch (
    mysqli_sql_exception $e) {
        print
    "Error Code <br>".$e->getCode();
        print
    "Error Message <br>".$e->getMessage();
        print
    "Strack Trace <br>".nl2br($e->getTraceAsString());
    }
    ?>
    Will print out something like
    Error Code: 0
    Error Message
    No index used in query/prepared statement select sess_value from frame_sessions where sess_name = '5b85upjqkitjsostvs6g9rkul1'
    Strack Trace:
    #0 G:classfileslib5database.php(214): mysqli->query('select sess_val...')
    #1 G:classfileslib5Session.php(52): database->getString('select sess_val...')
    #2 [internal function]: sess_read('5b85upjqkitjsos...')
    #3 G:classfilesincludes.php(50): session_start()
    #4 G:testsall_tests.php(4): include('G:dataWorkSit...')
    #5 {main}

    Anonymous

    3 years ago


    mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
    try {
                    $this->connection = mysqli_connect($hostname,$username,$password, $dbname);
    } catch (Exception $e) {
                    echo "Errno: " . mysqli_connect_errno() . PHP_EOL;
                    echo "Text error: " . mysqli_connect_error() . PHP_EOL;
                    exit;
    }

    За последние 24 часа нас посетили 11560 программистов и 827 роботов. Сейчас ищут 398 программистов …

    mysqli::$error

    mysqli_error

    (PHP 5, PHP 7)

    mysqli::$errormysqli_errorВозвращает строку с описанием последней ошибки

    Описание

    Объектно-ориентированный стиль

    Процедурный стиль

    string mysqli_error
    ( mysqli $link
    )

    Возвращаемые значения

    Строка с описанием ошибки. Пустая строка, если ошибки нет.

    Примеры

    Пример #1 Пример с $mysqli->error

    Объектно-ориентированный стиль


    <?php
    $mysqli 
    = new mysqli("localhost""my_user""my_password""world");/* check connection */
    if ($mysqli->connect_errno) {
        
    printf("Connect failed: %sn"$mysqli->connect_error);
        exit();
    }

    if (!

    $mysqli->query("SET a=1")) {
        
    printf("Errormessage: %sn"$mysqli->error);
    }
    /* close connection */
    $mysqli->close();
    ?>

    Процедурный стиль


    <?php
    $link 
    mysqli_connect("localhost""my_user""my_password""world");/* check connection */
    if (mysqli_connect_errno()) {
        
    printf("Connect failed: %sn"mysqli_connect_error());
        exit();
    }

    if (!

    mysqli_query($link"SET a=1")) {
        
    printf("Errormessage: %sn"mysqli_error($link));
    }
    /* close connection */
    mysqli_close($link);
    ?>

    Результат выполнения данных примеров:

    Errormessage: Unknown system variable 'a'
    

    Смотрите также

    • mysqli_connect_errno() — Возвращает код ошибки последней попытки соединения
    • mysqli_connect_error() — Возвращает описание последней ошибки подключения
    • mysqli_errno() — Возвращает код ошибки последнего вызова функции
    • mysqli_sqlstate() — Возвращает код состояния SQLSTATE последней MySQL операции

    Вернуться к: mysqli

    Правильная обработка ошибок позволяет сделать стабильное приложение, которое не будет завершаться неожиданно. В случае возникновения непредвиденной ситуации программа должна выдавать вменяемый ответ, почему она не хочет исполнится. И создание правильного механизма обработки ошибок — это задача программиста.

    В этой статье рассмотрим два вида ошибок, которые могут возникнуть при программировании взаимодействия с базой данных. Первый тип — это ошибка подключения к базе данных. Второй тип — это ошибка выполнения запроса к базе данных. Для обработки этих ошибок будем использовать специальные функции для работы с базой.

    Ошибка соединения с базой данных

    Сразу приведём пример обработки ошибки с соединением с базой данных:

    <?php 
       $host = 'localhost'; // адрес сервера
       $db_name = 'database'; // имя базы данных
       $user = 'user'; // имя пользователя
       $password = 'password'; // пароль
    
       // создание подключения к базе   
          $connection = mysqli_connect($host, $user, $password, $db_name);
    
       // проверка правильности подключения
          if(!$connection){ // при соединении с базой данных возникла ошибка
             echo 'Ошибка соединения: ' . mysqli_connect_error() . '<br>';
             echo 'Код ошибки: ' . mysqli_connect_errno();
          }else{ // соединение было установлено успешно
             // здесь можно делать запрос к базе, 
             // потому что соединение успешно установлено
          }
    ?>

    В этом примере можно заметить функцию mysqli_connect_error. Она выводит текстовое описание ошибки подключения (на английском языке). В отличии от неё функция mysqli_connect_errno выводит числовой код ошибки, к примеру «1045».

    Ошибка запроса к базе

    Теперь попробуем доработать пример из предыдущего параграфа и добавить запрос к базе данных. Не будем вдаваться в детали, что будет записано в тексте SQL запроса, просто предположим, что он может завершиться ошибкой. К примеру, из-за неожиданного отключения сервера с базой данных от сети. В примере добавим проверку на наличие ошибок при выполнении запроса:

    <?php 
       $host = 'localhost'; // адрес сервера
       $db_name = 'database'; // имя базы данных
       $user = 'user'; // имя пользователя
       $password = 'password'; // пароль
    
       // создание подключения к базе   
          $connection = mysqli_connect($host, $user, $password, $db_name);
    
          if(!$connection){ // проверка правильности подключения
             echo 'Ошибка соединения: ' . mysqli_connect_error() . '<br>';
             echo 'Код ошибки: ' . mysqli_connect_errno();
          }else{ // подключение успешно установлено
    
             // текст SQL запроса, который будет передан базе
                $query = 'SELECT * FROM `USERS`';
    
             // выполняем запрос к базе данных
                $result = mysqli_query($connection, $query);
    
                if(!$result){ // запрос завершился ошибкой
                   echo 'Ошибка запроса: ' . mysqli_error($connection) . '<br>';
                   echo 'Код ошибки: ' . mysqli_errno($connection);
                }else{ // запрос успешно выполнился
                   while($row = $result->fetch_assoc()){
                      // обрабатываем полученные данные
                   }
                }
             // закрываем соединение с базой
                mysqli_close($connection);
          }
    ?>

    В этом примере есть две функции, которые работают с ошибками базы. Функция mysqli_error возвращает описание ошибки запроса (на английском языке), а функция mysqli_errno возвращает числовой код ошибки, к примеру, «1193».

    Обратите внимание, что все функции обработки ошибок в этой статье (mysqli_connect_error, mysqli_connect_errno, mysqli_error, mysqli_errno) возвращают информацию только о последней ошибке. Но ошибок может быть несколько.

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    Last update on August 19 2022 21:50:40 (UTC/GMT +8 hours)

    mysqli_error() function / mysqli::$error

    The mysqli_error() function / mysqli::$error returns the last error description for the most recent function call, if any.

    Syntax:

    Object oriented style

    string $mysqli->error;
    

    Procedural style

    string mysqli_error ( mysqli $link )
    

    Parameter:

    Name Description Required/Optional
    link A link identifier returned by mysqli_connect() or mysqli_init() Required for procedural style only and Optional for Object oriented style

    Usage: Procedural style

    mysqli_error(connection);

    Parameter:

    Name Description Required/Optional
    connection Specifies the MySQL connection to use. Required

    Return value:

    A string that describes the error. An empty string if no error occurred.

    Version: PHP 5, PHP 7

    Example of object oriented style:

    <?php
    $mysqli = new mysqli("localhost", "user1", "datasoft123", "hr");
    
    /* check connection */
    if ($mysqli->connect_errno) {
        printf("Connect failed: %sn", $mysqli->connect_error);
        exit();
    }
    
    if (!$mysqli->query("SET a=1")) {
        printf("Errormessage: %sn", $mysqli->error);
    }
    
    /* close connection */
    $mysqli->close();
    ?>
    

    Output:

    Errormessage: Unknown system variable 'a'
    

    Example of procedural style:

    <?php
    
    $link = mysqli_connect("localhost", "my_user", "my_password", "world");
    
    /* check connection */
    if (mysqli_connect_errno()) {
        printf("Connect failed: %sn", mysqli_connect_error());
        exit();
    }
    
    if (!mysqli_query($link, "SET a=1")) {
        printf("Errormessage: %sn", mysqli_error($link));
    }
    
    /* close connection */
    mysqli_close($link);
    ?>
    

    Output:

    Errormessage: Unknown system variable 'a'
    

    Example:

    <?php
    $con=mysqli_connect("localhost","user1","datasoft123","hr");
    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
    
    // Perform a query, check for error
    if (!mysqli_query($con,"INSERT INTO employees (First_Name) VALUES ('David')"))
      {
      echo("Errorcode: " . mysqli_errno($con));
      }
    
    mysqli_close($con);
    ?>
    

    Sample Output:

    Errorcode: 1146
    

    See also

    PHP Function Reference

    Previous: error_list

    Next: field_count

    PHP: Tips of the Day

    PHP: How to Sort a Multidimensional Array in PHP?

    You can use array_multisort()

    Try something like this:

    foreach ($mdarray as $key => $row) {
        // replace 0 with the field's index/key
        $dates[$key]  = $row[0];
    }
    
    array_multisort($dates, SORT_DESC, $mdarray);
    

    For PHP >= 5.5.0 just extract the column to sort by. No need for the loop:

    array_multisort(array_column($mdarray, 0), SORT_DESC, $mdarray);

    Ref : https://bit.ly/2WkurVe

    Improve Article

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    The mysqli_error() function is used to return the error in the most recent MySQL function call that failed. If there are multiple MySQL function calls, the error in the last statement is the one that is pointed out by the function.

    Syntax:

    mysqli_error("database_name")
    

    Parameters: This function accepts single parameter as mentioned above and described below:

    • database_name: It is the database on which operations are being performed. It is a mandatory parameter.

    Program 1:

    <?php

    $conn = mysqli_connect(

        "localhost", "root", "", "Persons"); 

    if (mysqli_connect_errno()) { 

        echo "Database connection failed."

    if (!mysqli_query($link, "SET Age=1")) {

        printf("Error message: %sn", mysqli_error($conn));

    }    

    mysqli_close($conn);

    ?>

    Suppose the operation is being carried out on the table given below:

    The output will be:

    Error message: Unknown system variable 'Age'
    

    Program 2:

    <?php

    $conn = mysqli_connect(

        "localhost", "root", "", "Persons"); 

    if (mysqli_connect_errno()) { 

        echo "Database connection failed."

    if (!mysqli_query($link, "SET Firstname='Arkadyuti'")) {

        printf("Error message: %sn", mysqli_error());

    }    

    mysqli_close($conn);

    ?>

    Output:

    Error message: mysqli_error() expects exactly 1 parameter, 0 given
    

    This example also demonstrates that mysqli_error() needs a database as a parameter.

    Really, it drive me crazy when I can’t detect what error that happens. I can’t handle it.

    I managed to make a connection to MySQL, and check it out with:

    $connection = mysqli_connect(HOST, USER, PASS, DB) or die('Could not connect!');
    if($connection){
        echo 'It's connected!';
    }
    

    Yeah, that say connected. Then, when I try a query, it fails without error reporting. I’ve tried do this to check if it fails:

    $query = "SELECT $field FROM users WHERE id = ".$_SESSION['user_id'];
    $result = mysqli_query($dbc, $query);
    if($result){
        echo 'Query OK';
    }else{
        echo 'Query failed';
    }
    

    The browser said: Query failed. So, there’s an error in my query. Then I echoed the query out with this:

    echo $query;
    // Printed in the browser: SELECT firstname FROM users WHERE id = 1
    

    Copy that value and use it in phpMyAdmin. It works. So, i guess an error occured in mysqli_query function. But i can’t get the error message and so i don’t know what’s going on. I’ve tried this:

    $result = mysqli_query($dbc, $query) or die(mysqli_error($dbc));
    

    and this:

    if(!$result){
         echo mysqli_error($dbc);
    }
    

    Nothing happens. The browser just blank. Then, I tried to change this:

    $result = mysqli_query($dbc, $query);
    

    to this:

    $result = mysqli_query($query);
    

    Still nothing happens. What’s going on? How can I know what error occured?

    I run the server in Debian with phpinfo(): PHP Version 5.4.36-0+deb7u3

    1. PHP MySQLi Error Functions
    2. Conclusion

    Display Errors Using MySQLi Error Functions

    MySQLi is a PHP function used to access the MySQL database server. You can use this extension if you have MySQL version 4.1.13 or above.

    There are various MySQLi functions that you can use to perform different functions in PHP. In this article, we will learn MySQLi error functions.

    We will also see how and where to use code examples and observe the outputs. For this tutorial, we will use MySQL version 8.0.27 and PHP version 7.4.1.

    PHP MySQLi Error Functions

    In this tutorial, we will learn about the following PHP MySQLi error functions:

    • mysqli_error()
    • mysqli_errno()
    • mysqli_error_list()
    • mysqli_connect_error()
    • mysqli_connect_errno()

    All these functions can be used in object-oriented style and procedural style. Let’s understand both syntax using the mysqli_error() function.

    Syntax of mysqli_error() Function in Object Oriented Style

    Syntax of mysqli_error() Function in Procedural Style

    string mysqli_error ( mysqli $link )
    

    mysqli_error() Function in MySQL

    This function is used to output the last error description for the most recent function call (if there is any). It is used when you want to know whether the SQL query has an error or not.

    mysqli_error() returns the error description and empty string if there is no error. See the following example.

    Example Code Using Procedural Style:

    <?php
        $host = "localhost";
        $username = "root";
        $password = "";
        $database = "person";
    
        $connection = mysqli_connect($host, $username, $password, $database) 
        or die("Connection Failed"); 
    
        $sql = "SELECT * FROM teacher";
        $result = mysqli_query($connection, $sql);
        $error_message = mysqli_error($connection);
    
        if($error_message == ""){
            echo "No error related to SQL query.";
        }else{
            echo "Query Failed: ".$error_message;
        }
        mysqli_close($connection);
    ?>
    

    The code given above tries to make the connection using $host, $username, $password, $database variables and save this connection into the $connection variable.

    mysqli_error() function will take this connection variable $connection as a parameter and check if there is any error caused by the recent MySQLi function call which is mysqli_query($connection, $sql) here.

    Output:

    display errors using mysqli error functions - no error using mysqli_error

    Now, change the table name in the SQL query from teacher to person and observe the output given below.

    Output:

    display errors using mysqli error functions - error using mysqli_error

    We, as a developer, can easily understand that there is no person table in the person database (this is what it means in the above error).

    Keep the table name changed and replace the line $error_message = mysqli_error($connection); with $error_message = $connection->error; to practice and understand the object oriented style using MySQLi error function.

    mysqli_errno() Function in MySQL

    mysqli_errno() works the same as mysqli_error() does, but it will return the error code instead of the error description.

    Write the following code to practice and understand. You may have noticed that we use a procedural style to practice this function.

    <?php
         $host = "localhost";
         $username = "root";
         $password = "";
         $database = "person";
    
         $connection = mysqli_connect($host, $username, $password, $database) 
         or die("Connection Failed"); 
    
         $sql = "SELECT * FROM person";
         $result = mysqli_query($connection, $sql);
         $error_message = mysqli_errno($connection);
    
         if($error_message == ""){
         	echo "No error related to SQL query.";
         }else{
         	echo "Query Failed: ".$error_message;
         }
        mysqli_close($connection);
    ?>
    

    The code given above will show the following output where you will see a number as an error code.

    Output:

    display errors using msyqli error functions - error using mysqli_errno

    The question is, why do we use this function to show the numbers only? Because if you want to print a user-friendly error message (custom message), you can use this error code in if-else statements.

    See the following code and its output below.

    <?php
         $host = "localhost";
         $username = "root";
         $password = "";
         $database = "person";
    
         $connection = mysqli_connect($host, $username, $password, $database) 
         or die("Connection Failed"); 
    
         $sql = "SELECT * FROM person";
         $result = mysqli_query($connection, $sql);
         $error_message = mysqli_errno($connection);
    
         if($error_message == 1146){
         	echo "You are trying to read the data from a table which doesn't exist in your 			database "."'".$database."'";
         }
        mysqli_close($connection);
    ?>
    

    Output:

    display errors using mysqli error functions - custom error message using mysqli_errno.png

    mysqli_error_list() Function in MySQL

    This function is very useful for knowing the error code, SQL state, and error description because this function returns an array containing all the necessary information.

    Example Code:

    <?php
         $host = "localhost";
         $username = "root";
         $password = "";
         $database = "person";
    
         $connection = mysqli_connect($host, $username, $password, $database) 
         or die("Connection Failed"); 
    
         $sql = "SELECT * FROM person";
         $result = mysqli_query($connection, $sql);
         print_r(mysqli_error_list($connection));
         mysqli_close($connection);
    ?>
    

    Output:

    display errors using mysqli error functions - error list using mysqli_error_list

    mysqli_connect_error() Function in MySQL

    mysqli_connect_error() returns the error description from the last connection if there is any. Although, the die() function also tell about the unsuccessful connection but mysqli_connect_error() returns the error that we can understand easily.

    Write the following code first, see its output, and then we’ll compare it with the output produced by mysqli_connect_error().

    <?php
         $host = "localhost";
         $username = "root";
         $password = "";
         $database = "person";
    
         $connection = mysqli_connect($host, $username, $password, $database) 
         or die("Connection Failed"); 
    
         $sql = "SELECT * FROM person";
         $result = mysqli_query($connection, $sql);
         $error_message = mysqli_error($connection);
    
         if($error_message != ""){
         	echo "Query Failed: ".$error_message;
          }
        mysqli_close($connection);
    ?>
    

    Output:

    display errors using mysqli error functions - error using mysqli_connect_error part a

    See the output given above; you can see that the error we can understand is somewhere in the middle.

    Imagine, if you have 2 or 3 errors, it would not be easy to find out. Now, use the mysqli_connect_error() and see the difference using the following code and output.

    <?php
         $host = "localhost";
         $username = "newroot";
         $password = "";
         $database = "person";
    
         $connection = mysqli_connect($host, $username, $password, $database) 
         or die("Connection Failed: ".mysqli_connect_error()); 
    
         $sql = "SELECT * FROM teacher";
         $result = mysqli_query($connection, $sql);
         $error_message = mysqli_error($connection);
    
         if($error_message != ""){
         	echo "SQL Query Failed: ".$error_message;
        }
        mysqli_close($connection);
    ?>
    

    Output:

    display errors using mysqli error functions - error using mysqli_connect_error part b

    The above output clearly says that there is no user named newroot, which does not allow you to access the database.

    mysqli_connect_errno() Function in MySQL

    This function behaves like mysqli_connect_error() but displays the error code rather than the error message. We can use this error code to write custom error messages.

    Example Code:

    <?php
         $host = "localhost";
         $username = "newroot";
         $password = "";
         $database = "person";
    
         $connection = mysqli_connect($host, $username, $password, $database) 
         or die("Connection Failed: ".mysqli_connect_errno()); 
    
         $sql = "SELECT * FROM teacher";
         $result = mysqli_query($connection, $sql);
         $error_message = mysqli_error($connection);
    
         if($error_message != ""){
    	     echo "SQL Query Failed: ".$error_message;
         }
         mysqli_close($connection);
    ?>
    

    Output:

    display errors using mysqli error functions - error using mysqli_connect_errno

    Conclusion

    Considering all the discussion and examples, we have concluded two main categories. The first category shows the errors about SQL queries and the other about database connections.

    Depending on the project needs, we can print the error message or the error code in each category.

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