PHP - Scope of assignment within if statement

Question

How do I delcare a variable in an if statment that only goes in the if block if the variable is not a nullish value?

Also, what is the scope of the $var in code like:

if(($var = funky()) != null) {}

Can I reference $var outside the if block?

if($var = funky()) - I take it that that is pseudo code? If not, then what you're doing is assigning = rather than comparing ==, unless that's what you want to do. — Funk Forty Niner
– Funk Forty Niner, Commented May 3, 2017 at 16:33
"Will the code inside the if block still execute or do I need a conditional as well?" - Yes, because it will always be considered as TRUE, since it's an assignment. Your edit: "such as if(!empty($var = funky()))" - only if it's not empty. — Funk Forty Niner
– Funk Forty Niner, Commented May 3, 2017 at 16:34
it won't always be true, it'll assign the return value of the function, and that'll be evaluated — billyonecan
– billyonecan, Commented May 3, 2017 at 16:36
...which is what I said, just explained differently, for the latter part of the above comment. — Funk Forty Niner
– Funk Forty Niner, Commented May 3, 2017 at 16:36
@Fred-ii- i was referencing this: "Will the code inside the if block still execute or do I need a conditional as well?" - Yes, because it will always be considered as TRUE, since it's an assignment. < it won't if the function doesn't return truthy — billyonecan
– billyonecan, Commented May 3, 2017 at 16:49

Don't Panic · Accepted Answer · 2017-05-03 17:19:02Z

2

An assignment expression in PHP returns the assigned value. From the documentation:

The value of an assignment expression is the value assigned.

So if whatever funky() returns evaluates to something loosely equal to null, $var = funky() will evaluate to false, so the if block will not be executed.

Doing the assignment in the if condition does not affect the scope of the assigned variable, though. $var will be available in the current scope (inside and outside the if block) after the assignment statement.

For example:

function funky() {
    return false;
}

if(($var = funky()) != null) {
    echo 'something';
}

var_dump($var);

Here, you'll just see boolean false

The only way I can think of to ensure that a variable is only available inside an if block is to assign it there and unset it before the end of the block.

if (funky() != null) {  // evaluate funky() without assigning
    $var = funky();     // call funky() again to assign
    // do stuff
    unset($var);
}

But I can't really think of any good reason to do this.

edited May 3, 2017 at 17:19

answered May 3, 2017 at 16:43

Don't Panic

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3 Comments

Kellen Stuart Over a year ago

Wow. I am surprised $var is available outside the if block

billyonecan Over a year ago

@KolobCanyon why? You haven't altered the scope at all (an if statement isn't a closure). You've basically just assigned a variable, and then wrote an if statement, except you've done it all in the same line

Kellen Stuart Over a year ago

@billyonecan Gotcha. Did not know that. I always had tried to assign the variable and then do the if statement on the next line, but I saw some code that one lined it... which lead me to this question

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