Php warning create default object from empty ошибка

asked Jan 17, 2012 at 19:43

Your new environment may have E_STRICT warnings enabled in error_reporting for PHP versions <= 5.3.x, or simply have error_reporting set to at least E_WARNING with PHP versions >= 5.4. That error is triggered when $res is NULL or not yet initialized:

$res = NULL;
$res->success = false; // Warning: Creating default object from empty value

PHP will report a different error message if $res is already initialized to some value but is not an object:

$res = 33;
$res->success = false; // Warning: Attempt to assign property of non-object

In order to comply with E_STRICT standards prior to PHP 5.4, or the normal E_WARNING error level in PHP >= 5.4, assuming you are trying to create a generic object and assign the property success, you need to declare $res as an object of stdClass in the global namespace:

$res = new stdClass();
$res->success = false;

answered Jan 17, 2012 at 19:45

This message has been E_STRICT for PHP <= 5.3. Since PHP 5.4, it was unluckilly changed to E_WARNING. Since E_WARNING messages are useful, you don’t want to disable them completely.

To get rid of this warning, you must use this code:

if (!isset($res)) 
    $res = new stdClass();

$res->success = false;

This is fully equivalent replacement. It assures exactly the same thing which PHP is silently doing — unfortunatelly with warning now — implicit object creation. You should always check if the object already exists, unless you are absolutely sure that it doesn’t. The code provided by Michael is no good in general, because in some contexts the object might sometimes be already defined at the same place in code, depending on circumstances.

answered Jan 8, 2014 at 20:10

Simply,

    $res = (object)array("success"=>false); // $res->success = bool(false);

Or you could instantiate classes with:

    $res = (object)array(); // object(stdClass) -> recommended

    $res = (object)[];      // object(stdClass) -> works too

    $res = new stdClass(); // object(stdClass) -> old method

and fill values with:

    $res->success = !!0;     // bool(false)

    $res->success = false;   // bool(false)

    $res->success = (bool)0; // bool(false)

More infos:
https://www.php.net/manual/en/language.types.object.php#language.types.object.casting

answered Jan 17, 2016 at 18:25

If you put «@» character begin of the line then PHP doesn’t show any warning/notice for this line. For example:

$unknownVar[$someStringVariable]->totalcall = 10; // shows a warning message that contains: Creating default object from empty value

For preventing this warning for this line you must put «@» character begin of the line like this:

@$unknownVar[$someStringVariable]->totalcall += 10; // no problem. created a stdClass object that name is $unknownVar[$someStringVariable] and created a properti that name is totalcall, and it's default value is 0.
$unknownVar[$someStringVariable]->totalcall += 10; // you don't need to @ character anymore.
echo $unknownVar[$someStringVariable]->totalcall; // 20

I’m using this trick when developing. I don’t like disable all warning messages becouse if you don’t handle warnings correctly then they will become a big error in future.

answered Jul 4, 2017 at 12:59

Try this if you have array and add objects to it.

$product_details = array();

foreach ($products_in_store as $key => $objects) {
  $product_details[$key] = new stdClass(); //the magic
  $product_details[$key]->product_id = $objects->id; 
   //see new object member created on the fly without warning.
}

This sends ARRAY of Objects for later use~!

answered Jun 12, 2015 at 9:00

1. First think you should create object
   $res = new stdClass();

2. then assign object with key and value thay
   $res->success = false;

answered Apr 28, 2020 at 4:25

Try this:

ini_set('error_reporting', E_STRICT);

answered Sep 5, 2013 at 8:54

answered Apr 12, 2022 at 13:30

A simple way to get this error is to type (a) below, meaning to type (b)

(a) $this->my->variable

(b) $this->my_variable

Trivial, but very easily overlooked and hard to spot if you are not looking for it.

Anton PelykhAnton Pelykh

2,2461 gold badge18 silver badges21 bronze badges

Try using:

$user = (object) null;

answered Nov 8, 2018 at 22:10

AndyAndy

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answered Jan 17, 2012 at 19:46

This problem is caused because your are assigning to an instance of object which is not initiated. For eg:

Your case:

$user->email = 'exy@gmail.com';

Solution:

$user = new User;
$user->email = 'exy@gmail.com';

answered Jun 8, 2014 at 4:46

I put the following at the top of the faulting PHP file and the error was no longer display:

error_reporting(E_ERROR | E_PARSE);

answered Feb 6, 2016 at 14:59