CODEWITHSUNDEEP
phpvariablesscopevariable-assignment
Kellen Stuart
Kellen Stuart

Reputation: 8893

PHP - Scope of assignment within if statement

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?

Upvotes: 2

Views: 438

Answers (1)

Don't Panic
Don't Panic

Reputation: 41810

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.

Upvotes: 2

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