The variable is undefined ошибка

Today, I have started to learn PHP. And, I have created my first PHP file to test different variables. You can see my file as follows.

<?php
    $x = 5; // Global scope

    function myTest()
    {
        $y = 10; // Local scope
        echo "<p>Test variables inside the function:<p>";
        echo "Variable x is: $x";
        echo "<br>";
        echo "Variable y is: $y";
    }

    myTest();

    echo "<p>Test variables outside the function:<p>";
    echo "Variable x is: $x";
    echo "<br>";
    echo "Variable y is: $y";
?>

I have found the following errors when I have run this file in the browser.

Notice: Undefined variable: x in /opt/lampp/htdocs/anand/php/index.php on line 19

Notice: Undefined variable: y in /opt/lampp/htdocs/anand/php/index.php on line 29

How can I fix the issue regarding it?

Community's user avatar

asked Dec 5, 2013 at 4:59

Anand Mistry's user avatar

1

The first error ($x is undefined) is because globals are not imported into functions by default (as opposed to «super globals», which are).

You need to tell your function you’re referencing the global variable $x:

function myTest()
{
  global $x; // $x refers to the global variable

  $y=10; // local scope
  echo "<p>Test variables inside the function:<p>";
  echo "Variable x is: $x";
  echo "<br>";
  echo "Variable y is: $y";
}

Otherwise, PHP cannot tell whether you are shadowing the global variable with a local variable of the same name.

The second error ($y is undefined), is because local scope is just that, local. The whole point of it is that $y doesn’t «leak» out of the function. Of course you cannot access $y later in your code, outside the function in which it is defined. If you could, it would be no different than a global.

answered Dec 5, 2013 at 5:01

user229044's user avatar

user229044user229044

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<?php
    $a = 1; /* Global scope */

    function test()
    {
        echo $a; /* Reference to local scope variable */
    }

    test();
?>

You are getting the first error because the variable $a can’t access the global variable’s value unless you explicitly declare global $a inside the function.

Example #1 Using a global

<?php
    $a = 1;
    $b = 2;

    function Sum()
    {
        global $a, $b; // If you want to access a global variable,
                       // you have to use the 'global' keyword

        $b = $a + $b;
    }

    Sum();
    echo $b;
?>

And the last error you are getting because $y is defined inside the function mytest() so its scope will be limited to that function only.

For a detailed explanation, read Variable scope.

Peter Mortensen's user avatar

answered Dec 5, 2013 at 5:01

R R's user avatar

R RR R

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Set $x as a global, like

global $x;

Or try this:

<?php
    $x = 5; // Global scope

    function myTest($x)
    {
        $y=10; // Local scope
        echo "<p>Test variables inside the function:<p>";
        echo "Variable x is: $x";
        echo "<br>";
        echo "Variable y is: $y";
    }

    myTest($x);

    echo "<p>Test variables outside the function:<p>";
    echo "Variable x is: $x";
    echo "<br>";
    echo "Variable y is: $y";
?>

Peter Mortensen's user avatar

answered Dec 5, 2013 at 5:03

himanshu bhardiya's user avatar

There are two cases of using a variable globally:

  1. Using a single copy of that variable and modify it from anywhere i.e. modification from within a function or outside i.e. in the global scope. In that case you need a declaration in the allowed set of function in the form global $x;.
  2. In case you need local variables for individual functions with the same identifier used for the global variable (i.e. variables outside all functions); in that case you have two variables with the same name i.e. one local and one global for that function. Then you need to use a superglobal variable $GLOBALS i.e. an array of all the global variables. I personally prefer this approach to make efficient code;

The following are the code for the two.

Code 1 (using global declaration)

<?php
    $x = 5; // Global scope

    function myTest()
    {
        $y = 10; // Local scope
        global $x;

        echo "<p>Test variables inside the function:<p>";
        echo "Variable x in global scope is: $x";
        echo "<br>";
        echo "Variable y is: $y";
    }

    myTest();

    echo "<p>Test variables outside the function:<p>";
    echo "Variable x is: $x";
    echo "<br>";
    echo "Variable y is: $y";
?>

Code 2 (using the $GLOBALS[] array)

<?php
    $x = 5; // Global scope

    function myTest()
    {
        $y = 10; // Local scope
        $x = 23;

        echo "<p>Test variables inside the function:<p>";
        echo "Variable x in global scope is: " . $GLOBALS['x'];
        echo "<br>Variable x in local scope is: $x";
        echo "<br>";
        echo "Variable y is: $y";
    }

    myTest();

    echo "<p>Test variables outside the function:<p>";
    echo "Variable x is: $x";
    echo "<br>";
    echo "Variable y is: $y";
?>

For REFERENCE.

Peter Mortensen's user avatar

answered Dec 5, 2013 at 5:09

Rajesh Paul's user avatar

Rajesh PaulRajesh Paul

6,7336 gold badges40 silver badges56 bronze badges

The code is behaving as expected, but if you want to use both the variables across the script use this:

<?php
    $x = 5; // Global scope

    function myTest(){
        global $x;
        global $y;
        $y = 10;

        echo "<p>Test variables inside the function:<p>";
        echo "Variable x is: $x";
        echo "<br>";
        echo "Variable y is: $y";
    }
    myTest();

    echo "<p>Test variables outside the function:<p>";
    echo "Variable x is: $x";
    echo "<br>";
    echo "Variable y is: $y";
?>

Peter Mortensen's user avatar

answered Dec 5, 2013 at 5:14

Vipin Kumar Soni's user avatar

In PHP, global variables must be declared global inside a function if they are going to be used in that function.

function myTest()
{
    $y = 10; // Local scope
    global $x;
    .....
}

By declaring $x global within the function, it will refer to the global version of the variable.

Peter Mortensen's user avatar

answered Dec 5, 2013 at 5:03

Linga's user avatar

LingaLinga

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You have to learn the scope of a variable in PHP. See Variable scope.

In your code, $x is a global one, so in order to access that variable inside the function use global $x; at the beginning of the function, that is,

function myTest()
{
    global $x;
    $y = 10; // Local scope

    echo "<p>Test variables inside the function:<p>";
    echo "Variable x is: $x";
    echo "<br>";
    echo "Variable y is: $y";
}

For $y, either you skip that output by checking isset($y) or else assign a default value at the global scope.

Peter Mortensen's user avatar

answered Dec 5, 2013 at 5:04

gvgvgvijayan's user avatar

gvgvgvijayangvgvgvijayan

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put $x outside the»» in code like echo» Variable $x is:».$x;

answered Sep 4, 2021 at 10:08

PRAKHAR MISHRA's user avatar

  John Mwaniki /   10 Dec 2021

This error, as it suggests, occurs when you try to use a variable that has not been defined in PHP.

Example 1

<?php
echo $name;
?>

Output

Notice: Undefined variable: name in /path/to/file/file.php on line 2.

Example 2

<?php
$num1 = 27;
$answer = $num1 + $num2;
echo $answer;
?>

Output

Notice: Undefined variable: num2 in /path/to/file/file.php on line 3.

In our above two examples, we have used a total of 4 variables which include $name, $num1, $num2, and $answer.

But out of all, only two ($name and $num2) have resulted in the «undefined variable» error. This is because we are trying to use them before defining them (ie. assigning values to them).

In example 1, we are trying to print/display the value of the variable $name, but we had not yet assigned any value to it.

In example 2, we are trying to add the value of $num1 to the value of $num2 and assign their sum to $answer. However, we have not set any value for $num2.

To check whether a variable has been set (ie. assigned a value), we use the in-built isset() PHP function.

Syntax

isset($variable)

We pass the variable name as the only argument to the function, where it returns true if the variable has been set, or false if the variable has not been set.

Example

<?php
//Example 1
$name = "Raju Rastogi";
if(isset($name)){
  $result = "The name is $name";
}
else{
  $result = "The name has not been set";
}
echo $result;
//Output: The name is Raju Rastogi
echo "<br>"


//Example 2
if(isset($profession)){
  $result = "My profession is $profession";
}
else{
  $result = "The profession has not been set";
}
echo $result;
//Output: The profession has not been set
?>

In our first example above, we have defined a variable ($name) by creating it and assigning it a value (Raju Rastogi) and thus the isset() function returns true.

In our second example, we had not defined our variable ($profession) before passing to the isset() function and thus the response is false.

The Fix for Undefined variable error

Here are some ways in which you can get rid of this error in your PHP program.

1. Define your variables before using them

Since the error originates from using a variable that you have not defined (assigned a value to), the best solution is to assign a value to the variable before using it anywhere in your program.

For instance, in our first example, the solution is to assign a value to the variable $name before printing it.

<?php
$name = "Farhan Qureshi";
echo $name;
//Output: Farhan Qureshi
?>

The above code works without any error.

2. Validating variables with isset() function

Another way to go about this is to validate whether the variables have been set before using them.

<?php
$num1 = 27;
if(isset($num1) && isset($num2)){
$answer = $num1 + $num2;
echo $answer;
}
?>

The addition operation will not take place because one of the required variables ($num2) has not been set.

<?php
$num1 = 27;
$num2 = 8;
if(isset($num1) && isset($num2)){
$answer = $num1 + $num2;
echo $answer;
}
//Oputput: 35
?>

The addition this time will take place because the two variables required have been set.

3. Setting the undefined variable to a default value

You can also check whether the variables have been defined using the isset() function and if not, assign them a default value.

For instance, you can set a blank «» value for variables expected to hold a string value, and a 0 for those values expect to hold a numeric value.

Example

<?php
$name = "John Doe";
$name = isset($name) ? $name : '';
$age= isset($age) ? $age: 0;
echo "My name is $name and I am $age yrs old.";
//Output: My name is John Doe and I am 0 yrs old.
?>

4. Disabling Notice error reporting

This is always a good practice to hide errors from the website visitor. In case these errors are visible on the website to the web viewers, then this solution will be very useful in hiding these errors.

This option does not prevent the error from happening, it just hides it.

Open the php.ini file in a text editor, and find the line below:

error_reporting = E_ALL

Replace it with:

error_reporting = E_ALL & ~E_NOTICE

Now the ‘NOTICE’ type of errors won’t be shown again. However, the other types of errors will be shown.

Another way of disabling these errors is to add the line of code below on the top of your PHP code.

error_reporting (E_ALL ^ E_NOTICE);

That’s all for this article. It’s my hope that it has helped you solve the error.

Почему возникает ошибка

Ошибка undefined variable появляется при попытке обратиться к не существующей (не объявленной ранее) переменной:

<?php
echo $text;

Если в настройках PHP включено отображение ошибок уровня E_NOTICE, то при запуске этого кода в браузер выведется ошибка:

Notice: Undefined variable: text in D:ProgramsOpenServerdomainstest.localindex.php on line 2

Как исправить ошибку

Нужно объявить переменную перед обращением к ней:

$a = '';
echo $a;

Нет уверенности, что переменная будет существовать? Можно указать значение по-умолчанию:

<?php
if(!isset($text))
    $text = '';

echo $text;

Или сокращённые варианты:

<?php
// С оператором объединения с null (PHP 7+)
$text = $text ?? '';

// С оператором присваивания значения null (PHP 7.4+)
$text ??= '';

Есть ещё один вариант исправления этой ошибки — отключить отображение ошибок уровня E_NOTICE:

<?php
error_reporting(E_ALL & ~E_NOTICE);
echo $a; // Ошибки не будет

Не рекомендую этот вариант. Скрытие ошибок вместо их исправления — не совсем правильный подход.

Кроме этого, начиная с PHP 8 ошибка undefined variable перестанет относиться к E_NOTICEи так легко отключить её уже не удастся.

Если ошибка появилась при смене хостинга

Часто ошибка возникает при переезде с одного сервера на другой. Практически всегда причина связана с разными настройками отображения ошибок на серверах.

По-умолчанию PHP не отображает ошибки уровня E_Notice, но многие хостинг-провайдеры предпочитают настраивать более строгий контроль ошибок. Т.е. на старом сервере ошибки уже были, но игнорировались сервером, а новый сервер таких вольностей не допускает.

Остались вопросы? Добро пожаловать в комментарии. :)

11.03.2021

Марат

17

0

php | php_error |

undefined variable или как исправить ошибку «undefined variable» -я даже и забыл, что такой вид ошибки существует!

Подробно об ошибке «undefined variable«

  1. undefined variable
  2. Как выключить ошибку undefined variable
  1. Notice undefined variable

    Что означает «Notice undefined variable» — начнем с перевода… Когда вы встречаете разного рода ошибки — просто переведите данную ошибку — потому, что ошибка она на то и ошибка, что в смысле этой ошибки и есть ответ.

    Notice — переводится как :

    уведомление, внимание, объявление, нотис сущ
    замечать, уведомлять, обращать внимание гл

    undefined — переводится как :

    неопределенный прил

    variable — переводится как :

    переменная сущ
    изменчивый прил
    регулируемый, изменяющийся прич

    Итого, если суммировать и перевести с русского на русский:

    «Notice undefined variable» — внимание! Неопределенная переменная…

    Т.е. вы выводите переменную, которая не существует! Вообще — это очень неудобная ошибка, я её просто выключил!

    Пример возникновения «undefined variable»

    Самый простой пример… возникновения ошибки «undefined variable«. Переменная ранее не была создана и выводим таким образом с помощью echo:

    echo $example_var;

    Для предотвращения такого рода ошибки, выше echo нужно создать данную переменную:

    $example_var = »;

    echo $example_var;

  2. Как выключить ошибку undefined variable

    Почему я написал выше, что я забыл, что вообще существует такой вид ошибки, как «» — да просто потому, что я каждый день не покупаю новый хостинг, а нахожусь уже на одном хостинге ruweb — 10 лет

    И у меня с самого начала, данная ошибка была отключена.

    Но если у вас не так.
    Стоит ли отключать вывод ошибки undefined variable.

    По этому поводу — всегда будут два мнения за и против.

    Я отключаю, потому, что мне не нужна именно эта ошибка.

    Смотри как отключить ошибки, или как показывать ошибки, только админу.

Не благодарите, но ссылкой можете поделиться!

COMMENTS+

 
BBcode


Notice: Undefined variable

This error means that within your code, there is a variable or constant which is not set. But you may be trying to use that variable.

The error can be avoided by using the isset() function.This function will check whether the variable is set or not.

Error Example:

<?php
$name='Stechies';
echo $name;
echo $age;
?>

Output:

STechies
Notice: Undefined variable: age in testsite.locvaraible.php on line 4

In the above example, we are displaying value stored in the ‘name’ and ‘age’ variable, but we didn’t set the ‘age’ variable.

Notice: Undefined offset error in PHP

Here are two ways to deal with such notices.

  1. Resolve such notices.
  2. Ignore such notices.

Fix Notice: Undefined Variable by using isset() Function

This notice occurs when you use any variable in your PHP code, which is not set.

Solutions:

To fix this type of error, you can define the variable as global and use the isset() function to check if the variable is set or not.

Example:

<?php
global $name;
global $age;
$name = 'Stechies';

if(!isset($name)){
$name = 'Variable name is not set';
}
if(!isset($age)){
$age = 'Varaible age is not set';
}
echo 'Name: ' . $name.'<br>';
echo 'Age: ' . $age;
?>

Set Index as blank

<?php
$name = 'Stechies';

// Set Variable as Blank
$name = isset($name) ? $name : '';
$age= isset($age) ? $age: '';

echo 'Name: ' . $name.'<br>';
echo 'Age: ' . $age;
?>

Ignore PHP Notice: Undefined variable

You can ignore this notice by disabling reporting of notice with option error_reporting.

1. Disable Display Notice in php.ini file

Open php.ini file in your favorite editor and search for text “error_reporting” the default value is E_ALL. You can change it to E_ALL & ~E_NOTICE.

By default:

error_reporting = E_ALL

Change it to:

error_reporting = E_ALL & ~E_NOTICE

Now your PHP compiler will show all errors except ‘Notice.’

2. Disable Display Notice in PHP Code

If you don’t have access to make changes in the php.ini file, In this case, you need to disable the notice by adding the following code on the top of your PHP page.

<?php error_reporting (E_ALL ^ E_NOTICE); ?>

Now your PHP compiler will show all errors except ‘Notice.’

By default:

Change it to:

Now your PHP compiler will show all errors except ‘Notice.’

2. Disable Display Notice in PHP Code

If you don’t have access to make changes in the php.ini file, In this case, you need to disable the notice by adding the following code on the top of your PHP page.

Now your PHP compiler will show all errors except ‘Notice.’

Источник

[Solved]: Notice: Undefined variable in PHP

This error, as it suggests, occurs when you try to use a variable that has not been defined in PHP.

Example 1

Notice: Undefined variable: name in /path/to/file/file.php on line 2.

Example 2

Notice: Undefined variable: num2 in /path/to/file/file.php on line 3.

In our above two examples, we have used a total of 4 variables which include $name , $num1 , $num2 , and $answer .

But out of all, only two ($name and $num2) have resulted in the «undefined variable» error. This is because we are trying to use them before defining them (ie. assigning values to them).

In example 1, we are trying to print/display the value of the variable $name, but we had not yet assigned any value to it.

In example 2, we are trying to add the value of $num1 to the value of $num2 and assign their sum to $answer. However, we have not set any value for $num2.

To check whether a variable has been set (ie. assigned a value), we use the in-built isset() PHP function.

Syntax

We pass the variable name as the only argument to the function, where it returns true if the variable has been set, or false if the variable has not been set.

Example

In our first example above, we have defined a variable ($name) by creating it and assigning it a value (Raju Rastogi) and thus the isset() function returns true.

In our second example, we had not defined our variable ($profession) before passing to the isset() function and thus the response is false.

The Fix for Undefined variable error

Here are some ways in which you can get rid of this error in your PHP program.

1. Define your variables before using them

Since the error originates from using a variable that you have not defined (assigned a value to), the best solution is to assign a value to the variable before using it anywhere in your program.

For instance, in our first example, the solution is to assign a value to the variable $name before printing it.

The above code works without any error.

2. Validating variables with isset() function

Another way to go about this is to validate whether the variables have been set before using them.

The addition operation will not take place because one of the required variables ($num2) has not been set.

The addition this time will take place because the two variables required have been set.

3. Setting the undefined variable to a default value

You can also check whether the variables have been defined using the isset() function and if not, assign them a default value.

For instance, you can set a blank «» value for variables expected to hold a string value, and a for those values expect to hold a numeric value.

Example

4. Disabling Notice error reporting

This is always a good practice to hide errors from the website visitor. In case these errors are visible on the website to the web viewers, then this solution will be very useful in hiding these errors.

This option does not prevent the error from happening, it just hides it.

Open the php.ini file in a text editor, and find the line below:

Replace it with:

Now the ‘NOTICE’ type of errors won’t be shown again. However, the other types of errors will be shown.

Another way of disabling these errors is to add the line of code below on the top of your PHP code.

error_reporting (E_ALL ^ E_NOTICE);

That’s all for this article. It’s my hope that it has helped you solve the error.

Источник

PHP: Ошибки при работе с переменными

Главное правило: переменная должна быть объявлена до её использования. Если сделать это позже, то программа просто не заработает.

Запуск программы выше завершается с ошибкой PHP Notice: Undefined variable: greeting. Undefined variable — это ошибка обращения. Она означает, что в коде используется имя (говорят идентификатор), который не определен. Причём в самой ошибке явно указан индентификатор переменой: greeting. Кроме неправильного порядка определения, в PHP встречаются банальные опечатки — как при использовании переменной, так и во время её объявления.

Количество подобных ошибок уменьшается за счёт использования правильно настроенного редактора. Такой редактор подсвечивает имена, которые используются без объявления, и предупреждает о возможных проблемах.

Задание

Найдите в программе необъявленную переменную и объявите её, присвоив ей значение ‘Dragon’ .

После выполнения программы результат на экране должен выглядеть так:

Если вы зашли в тупик, то самое время задать вопрос в «Обсуждениях». Как правильно задать вопрос:

  • Обязательно приложите вывод тестов, без него практически невозможно понять что не так, даже если вы покажете свой код. Программисты плохо исполняют код в голове, но по полученной ошибке почти всегда понятно, куда смотреть.

Тесты устроены таким образом, что они проверяют решение разными способами и на разных данных. Часто решение работает с одними входными данными, но не работает с другими. Чтобы разобраться с этим моментом, изучите вкладку «Тесты» и внимательно посмотрите на вывод ошибок, в котором есть подсказки.

Это нормально 🙆, в программировании одну задачу можно выполнить множеством способов. Если ваш код прошел проверку, то он соответствует условиям задачи.

В редких случаях бывает, что решение подогнано под тесты, но это видно сразу.

Создавать обучающие материалы, понятные для всех без исключения, довольно сложно. Мы очень стараемся, но всегда есть что улучшать. Если вы встретили материал, который вам непонятен, опишите проблему в «Обсуждениях». Идеально, если вы сформулируете непонятные моменты в виде вопросов. Обычно нам нужно несколько дней для внесения правок.

Кстати, вы тоже можете участвовать в улучшении курсов: внизу есть ссылка на исходный код уроков, который можно править прямо из браузера.

Источник

Troubleshooting Code 800A03F4 – Variable is undefined

Introduction to Code 800A03F4

Error code 800A03F4 occurs when you execute a VBScript.  To be blunt this is due to carelessness.  Either you have not declared a variable at all, or you have misspelt it later in the script.Code 800A03F4 Error: Variable Undefined

The Symptoms You Get

The script does not execute as you hoped, instead you get a WSH error message.

The Cause of Error 800A03F4

Code 800A03F4 means that you are using Option Explicit but you have not declared a variable as a Dim Statement in your script.  Note that the Source: is a runtime error (and not compilation error).

The Solution to Variable is undefined

The key to solving error 800A03F4 is reading the reference, ‘Error: Variable is undefined: ‘xyz’.  In my example ‘xyz’ is ‘numusers’.  You could also double check to see if you are using Option Explicit.  If so, then you need to declare the variable mentioned in the error message.  In my example, Windows Scripting Host is telling us that the problem is at Line: 239, in addition to regular lines, count blank lines and lines with remarks. 

Beware you may, as in my example have correctly declared a similar variable, and all this error is a typo, ‘numusers’ instead of ‘intNumusers’.   My point is that declaring ‘numusers’, would stop the error message but would result in a faulty script.  Preventing such problems is the reason I have Option Explicit alert me to potential logical errors.

In this case the Char: number (1), may not be the place on the line where the variable is found, so check the whole line.

  ‡

Example of Error 800A03F4 Script

You can see Dim statements at the beginning of the script, also observe that Option Explicit is set at the beginning of the script.

Option Explicit

‘ Objects
Dim objRootDSE, objTrans, objFSO, objShell
Dim objExcel, objSheet
Dim objUser, objGroup, objContainer

Dim intNumusers

 

Then go to line 239 and see the variable numusers, this is the root cause of the problem.  The solution is, either  declare this variable at the start or, as it is a typo, alter numusers to intNumusers

‘…..Cut to line line 231

231  End If
232  ‘ Increment to next user.
233  intRow = intRow + 1
234  intNumusers = intNumusers + 1
235

236  Loop
237
238  ‘ Popup message to finish
239  Wscript.Echo «Number of users created = » & numusers
 

See More Windows Update Error Codes 8004 Series

• Error 800A101A8 Object Required   •Error 800A0046   •Error 800A10AD   •Error 800A000D

• Error 80048820   •Error 800A0401   •Review of SolarWinds Permissions Monitor

• Error 80040E14   • Error 800A03EA   • Error 800A0408   • Error 800A03EE

Solarwinds Free WMI MonitorGuy Recommends: WMI Monitor and It’s Free!

Windows Management Instrumentation (WMI) is one of the hidden treasures of Microsoft operating systems.  Fortunately, SolarWinds have created the WMI Monitor so that you can examine these gems of performance information for free.  Take the guess work out of which WMI counters to use for applications like Microsoft Active Directory, SQL or Exchange Server.

Download your free copy of WMI Monitor


Do you need additional help?

  • For interpreting the WSH messages check Diagnose 800 errors.
  • For general advice try my 7 Troubleshooting techniques.
  • See master list of 0800 errors.
  • Codes beginning 08004…
  • Codes beginning 08005…
  • Codes beginning 08007…
  • Codes beginning 0800A…

Give something back?

Would you like to help others?  If you have a good example of this error, then please email me, I will publish it with a credit to you:

If you like this page then please share it with your friends


About The Author

Guy Recommends: WMI Monitor and It’s Free!

Windows Management Instrumentation (WMI) is one of the hidden treasures of Microsoft operating systems.  Fortunately, SolarWinds have created the WMI Monitor so that you can examine these gems of performance information for free.  Take the guess work out of which WMI counters to use for applications like Microsoft Active Directory, SQL or Exchange Server.

Download your free copy of WMI Monitor


Do you need additional help?

  • For interpreting the WSH messages check Diagnose 800 errors.
  • For general advice try my 7 Troubleshooting techniques.
  • See master list of 0800 errors.
  • Codes beginning 08004…
  • Codes beginning 08005…
  • Codes beginning 08007…
  • Codes beginning 0800A…

Give something back?

Would you like to help others?  If you have a good example of this error, then please email me, I will publish it with a credit to you:

If you like this page then please share it with your friends


About The Author

Guy Thomas

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03.10.2015, 18:01

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как можно проверить решение ?

1. Можно выбрать несколько кандидатов в решения и сравнить с полученным решением. Если они хуже найденного решения, то можно надеяться, что решение правильное. Но гарантии никакой.
2. Решение такой задачи существует, если область, где ищем решение, ограничена. И решение лежит на границе области (в данном случае — это многогранник). Это может быть 1) одна из вершин многогранника, 2) все точки какого то ребра многогранника, 3) все точки какой то грани. Это значит, что хотя бы один из элементов вектора A2*x совпадает с соответствующим элементом вектора b1 или b2. В вашем случае так и есть: А2*х0 = b20 = 6.4. Для гарантии возьмите еще две точки этой грани многогранника, не лежащие на одной прямой. Если результаты такие же, то на всей этой грани фукнционал достигает максимума.
3. Однако, вы изменили условия задачи (заменили неравенство на противоположное). Тогда ваша область — многогранник — ограничена только двумя плоскостями (двугранный угол). Это неограниченная область, в которой функция может не достичь максимума. Спросите, почему двугранный угол? Судите сами: Если выполнены два неравенства a<b, a<c, то одно из них лишнее и его можно отбросить. Например, если b<c, то оставляем первое неравенство a<b, так как второе автоматически выполняется.
Таким образом, теперь у вас в условии задачи на самом деле только два неравенства, определяющие двугранный угол.
4. Замечу, что четырех неравенств тоже может быть недостаточно для получения ограниченного многогранника
(для вашего (неисправленного) случая это я не проверял). Но составители задач обычно «подыгрывают» нам — решателям и подсовывают задачи, имеющие решения (если, конечно, не ошиблись).
5. Можно построить графики этих областей для наглядности.
6. Можно решить графическим методом такие задачи.
7. Можно, наконец, анимировать процесс решения.
8. Но все это в другой раз, если, конечно, интересно.



1



This guide will you how to solve the notice undefined variable, index, or offset error that you are experiencing in PHP. This error is easy to spot in the warning and error message logs. Consequently, you will typically see a descriptive error message like this:

1. Notice: Undefined variable
2. Notice: Undefined index
3. Notice: Undefined offset

Issue description

The notice undefined variable, index, and offset errors are very common in PHP. So, let’s look at some code examples that will output these errors.

Example 1

Here is an example of a code that will output a notice undefined variable error:

<?php
$name=’peter';

echo $name;
//undefined variable ‘age’
echo age;
?>

Notice that the code has an undefined age variable.

Example 2

Next, we have an example of code showing the error notice undefined index error:

<?php
$index = 1;
$file = $_POST['file'];

echo $file;
//undefined index ‘file’
?>

Notice that the $_POST array has an undefined “file” index.

Example 3

Finally, we have code showing undefined index and offset error in PHP:

<?php
// Declare and initialize an array
// $students = ['June', 'Abdi', 'Moha']
$students = array(
	0 => 'June',
	1 => 'Abdi',
	2 => 'Moha'
);

// June
echo $students[0];

// ERROR: Undefined offset: 4
echo $students[4];

// ERROR: Undefined index: num
echo $students[num];
?>

Possible causes

Now, let’s look at the possible causes of notice undefined variable, index, and offset errors.

Undefined variable

The undefined variable occurs due to one of the following reasons:

  • When you try to access an undefined variable in PHP.
  • Spelling mistakes: For example, if you define a variable $student but type in $students. These are two different variables that cannot be used interchangeably.
  • Not following the variable declaration rules in PHP.

Undefined offset

This error mostly appears when an offset does not appear in an array-like in our example 3 above.

Undefined index

This notice error mostly appears when you are trying to process variables not sent by a form.

For example, if you save a posted value using the following code:

$user_email = $_POST[‘user_email’];

The index error will not appear while submitting the form because $user_email can hold an empty string. However, when you access the page directly using an URL, PHP will display a ‘Notice Undefined Index’ error.

Possible solutions

Let’s discuss possible solutions for the undefined variable, index, and offset notice errors:

Follow all the variable declaration rules and classification

When you get an “undefined Variable” error message, you first need to define all variables.

In our example, we have defined both $name and $age to avoid the error. Next, check if your variables have spelling mistakes and correct them.

Note that variables are case-sensitive. Super global variables such as $_POST, $_GET, or $_SESSION should be uppercase.

Here are the rules to follow when declaring variables.

Variable declaration rules:

  • Each variable must start with a $ sign.
  • Ensure there is no space in the variable name.
  • The first letter should be derived from a-z, A-Z, or underscore _.
  • Variable names are case-sensitive. For example, $name and $NAME are two different variables.
  • The subsequent letters can be an underscore, a-z, A-Z, or 0-9.
  • The name cannot start with a letter.

There are two types of variables:

  • Predefined variables.
  • User-defined variables.

Predefined variables:

  1. $_GET
  2. $_POST
  3. $_COOKIE
  4. $_GLOBALS
  5. $_SESSION
  6. $_SERVER
  7. $_FILES
  8. $_REQUEST
  9. $_ENV
  10. $argc
  11. $argv
  12. $http_response_header

User-defined variables:

  1. Variable scope
  2. Variable variables
  3. Reference variables

If you are using a global variable, you need to check if an index exists. For example, you need to initialize $_SESSION with session_start(); otherwise, you will get an undefined index error. Once you follow all the declaration rules and the errors persist, you can try Fix 2 below.

Check if variables are defined before referencing them

When including one file into another, an uninitialized variable can be problematic. It can cause a huge security risk for turning on registered globals. This is why most PHP editors will display an E_NOTICE level error. You can use isset() to detect if the variable is initialized. Even better, you can use the empty() function to prevent the warning if a variable has not been initialized.

The isset() function checks whether you have set the variable. It also checks for null values in your array or array key. Additionally, the empty() function checks whether your array is empty.

The following examples prevent undefined offset error using isset() and empty() functions respectively

isset()

<?php
// Declare and initialize an array
// $students = ['June', 'Abdi', 'Moha']
$students = array(
	0 => 'June',
	1 => 'Abdi',
	2 => 'Moha'
);

if(isset($students[4])) {
  echo $students[4];
} else {
  echo "Index is not set";
}
?>

Output:

empty():

<?php
// Declare and initialize an array
// $students = ['June', 'Abdi', 'Moha']
$students = array(
	0 => 'June',
	1 => 'Abdi',
	2 => 'Moha'
);

if(!empty($students[4])) {
    echo $students[4];
} else {
    echo "Index is empty";
}
?>

Output:

Hide redundant or undesired error notices

Sometimes you can get redundant error notices, but you have to ensure your code is good before hiding the errors. You can hide undefined variables, index, and offset notices by using the @ error control operator to hide them.

For example:

 $var = @($_GET["optional_param"]);

This is highly discouraged because beginners tend to overuse it. Ignoring undeclared variables for loops and function parameters is very inappropriate.

Wrapping up

You can solve the undefined variable, index, and offset notices in many ways. First, if the variable is undefined, it simply means that you have not declared a variable. You need to follow all the guidelines when declaring the variable to avoid the error. Next, when transferring data to another page, you need to check if a variable is set to prevent the undefined index notice. Lastly, it would be best to ensure that the variables are assigned values to prevent undefined offset error messages. You can ignore redundant errors, but it’s not recommended.

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