I need to check if a variable is an object of the User type.
User is my class $user my object
$this->assertInstanceOf($user, User);
This is not working. I have a the following error: use of undefined constant User - assumed 'User'.
Add a comment
|
3 Answers
https://docs.phpunit.de/en/9.5/assertions.html#assertinstanceof
I think you are using this function wrong. Try:
$this->assertInstanceOf('User', $user);
As of PHP 5.5 you can also use:
$this->assertInstanceOf(User::class, $user);
(From @james2doyle in comments.)
Comments
It's always a good idea to use ::class wherever you can. If you get used to this standard, you don't have to use FQCNs (fully qualified classnames), or escape backslashes. Also, IDEs provide better functionality if they know that User here is not just a string, but rather a class.
$this->assertInstanceOf(User::class, $user);
4 Comments
rob74
I fully agree with this statement, but it sounds more like a comment on the accepted answer than like a separate answer...
Alex
I think this is the required way on PHP Unit 9.5.
Peter Mortensen
Why is it a good idea to use
::class?
Chuck Le Butt
@PeterMortensen Because it could be a string, as he says
Or you can use something like:
$this->assertInstanceOf(get_class($expectedObject), $user);
I usually use this when I'm checking i.e. if setter method is returning reference to self.
$testedObj = new ObjectToTest();
$this->assertInstanceOf(
get_class($testedObj),
$testedObj->setSomething('someValue'),
'Setter is not returning $this reference'
);
3 Comments
marcegarba
Using PHP 5.6, you could use the
::class static method, like this: $this->assertInstanceOf(ObjectToTest::class, $testedObj->setSomething('someValue')
james2doyle
Actually, the
SomeClass::class feature was added in PHP 5.5
Rafal Kozlowski
Yes, of course, but this is not a problem I guess. Services using php version 5.4 and lower are under 10% of total php usages (according to composer stats) seld.be/notes/php-versions-stats-2016-1-edition