PHP - Variable assignment in function call

Question

I have the following code where I try to assign a value to a variable and then evaluate if it's empty:

if(!empty($user = User::find($id))){
   // do some operations on the $user
}

But I'm getting an error... Does the PHP compiler allows this in any way?

PS. I'm using PHP 5.3

If you have $user previously in the context then put == instead of = otherwise the return value of User::find is not suitable for empty. — MahanGM
– MahanGM, Commented Dec 22, 2012 at 16:32
@MahanGM I believe he's assigning a value rather than comparing it — Adam Elsodaney
– Adam Elsodaney, Commented Dec 22, 2012 at 16:33
php is an interpreted language. what php compiler are you using, hiphop? can you provide some additional context? also, I think you should reconsider checking for !empty on the find method you are calling with an injected value. If you were unit testing this it would be a mess. break out your functionality into reuseable units. — jsteinmann
– jsteinmann, Commented Dec 22, 2012 at 16:39

dev-null-dweller · Accepted Answer · 2012-12-22 16:37:40Z

3

Does the PHP compiler allows this in any way?

Prior to PHP 5.5, empty() only supports variables; anything else will result in a parse error.

In your code you have assignment operation instead of plain variable.

answered Dec 22, 2012 at 16:37

dev-null-dweller

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Comments

Laurence · Accepted Answer · 2012-12-22 16:32:34Z

1

Generally if a function cant find something, ie a User, then it would return false anyway - so this is what I normally do:

if($user = User::find($id))
   // do some operations on the $user
}
else
{
   // no user found 
}

answered Dec 22, 2012 at 16:32

Laurence

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