PHP - Declare variable in if statement and use it in the same statement

Question

I want to declare a variable in a if statement and use it in the same statement. Is this simply not possible?

Example of problem (not actual use case though...):

if ($test = array('test'=>5) && $test['test'] === 5) {
    echo 'test';
} else {
    echo 'nope';
}

Error message:

NOTICE Undefined variable: test on line number 6

Operator precedence: if (($test = array('test'=>5)) && $test['test'] === 5) { echo 'test'; } else { echo 'nope'; } though I wouldn't consider this good coding practise — Mark Baker
– Mark Baker, Commented Nov 2, 2017 at 15:16
@MarkBaker Thanks! Why do you not consider this a good coding practise? — user7803115
– user7803115, Commented Nov 2, 2017 at 15:19
Why would you ever really need to do this? Just curious on the actual usability of this, I can't think of a scenario where it's useful. — GrumpyCrouton
– GrumpyCrouton, Commented Nov 2, 2017 at 15:22
Assigning variables inside if expressions can lead to confusion and errors. Someone else (or even you at a later point in time) can't be sure if it was meant to be an assignment or an invalid comparison. — M. Eriksson
– M. Eriksson, Commented Nov 2, 2017 at 15:23
This way I can check whether a record is successfully retrieved AND is the compatible with a previously retrieved record, all in one statement. Otherwise I would have to nest if statements. with each else having to do the exact same thing (404 not found) — user7803115
– user7803115, Commented Nov 3, 2017 at 10:31

AbraCadaver · Accepted Answer · 2021-01-27 15:31:12Z

9

Because of Operator Precedence you will need to group the assignment with ():

if ( ($test = array('test'=>5)) && $test['test'] === 5) {
    echo 'test';
} else {
    echo 'nope';
}

A simple use case might be:

if ( ($parts = parse_url($url)) && $parts['port'] == 8080) {
    // do stuff
}

answered Nov 2, 2017 at 15:16

AbraCadaver

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