Reputation: 11756
I see this error only after upgrading my PHP environment to PHP 5.4 and beyond. The error points to this line of code:
Error:
Creating default object from empty value
Code:
$res->success = false;
Do I first need to declare my $res
object?
Upvotes: 438
Views: 698465
Reputation: 104
First think you should create object $res = new \stdClass();
then assign object with key and value $res->success = false;
Upvotes: 2
Reputation: 31
Starting from PHP 7 you can use a null coalescing operator to create a object when the variable is null.
$res = $res ?? new \stdClass();
$res->success = false;
Upvotes: 3
Reputation: 7525
In PHP 7 anonymous objects can be created this way:
$res = new class {
public $success = false;
};
https://www.php.net/manual/en/language.oop5.anonymous.php http://sandbox.onlinephpfunctions.com/code/ab774707a8219c0f35bdba49cc84228b580b52ee
Upvotes: 4
Reputation: 2463
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
Upvotes: 19
Reputation: 115
This is a warning which I faced in PHP 7, the easy fix to this is by initializing the variable before using it
$myObj=new \stdClass();
Once you have intialized it then you can use it for objects
$myObj->mesg ="Welcome back - ".$c_user;
Upvotes: 0
Reputation: 343
I had a similar problem while trying to add a variable to an object returned from an API. I was iterating through the data with a foreach loop.
foreach ( $results as $data ) {
$data->direction = 0;
}
This threw the "Creating default object from empty value" Exception in Laravel.
I fixed it with a very small change.
foreach ( $results as &$data ) {
$data->direction = 0;
}
By simply making $data a reference.
I hope that helps somebody a it was annoying the hell out of me!
Upvotes: 0
Reputation: 270757
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;
Upvotes: 752
Reputation: 156
I had similar problem and this seem to solve the problem. You just need to initialize the $res object to a class . Suppose here the class name is test.
class test
{
//You can keep the class empty or declare your success variable here
}
$res = new test();
$res->success = false;
Upvotes: 1
Reputation: 2390
You may need to check if variable declared and has correct type.
if (!isset($res) || !is_object($res)) {
$res = new \stdClass();
// With php7 you also can create an object in several ways.
// Object that implements some interface.
$res = new class implements MyInterface {};
// Object that extends some object.
$res = new class extends MyClass {};
}
$res->success = true;
Upvotes: 0
Reputation: 65
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.
Upvotes: 0
Reputation: 995
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.
Upvotes: 8
Reputation: 5276
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);
Upvotes: -12
Reputation: 7723
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~!
Upvotes: 6
Reputation: 938
This problem is caused because your are assigning to an instance of object which is not initiated. For eg:
Your case:
$user->email = '[email protected]';
Solution:
$user = new User;
$user->email = '[email protected]';
Upvotes: -1
Reputation: 59575
This message has been E_STRICT
for PHP <= 5.3. Since PHP 5.4, it was unluckilly changed to E_WARNING
. Since E_WARNING
messages are useful, you don't want to disable them completely.
To get rid of this warning, you must use this code:
if (!isset($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.
Upvotes: 68
Reputation: 1680
no you do not .. it will create it when you add the success value to the object.the default class is inherited if you do not specify one.
Upvotes: -1