PHP offset warning on null

Question

I should be getting an offset warning error in this code but it shows nothing. What am i missing ?

I have turned on error reporting etc.

<?php
$info = null;
$value = 'string';
if ($value == $info['firstname']){
 echo 'true';
}else{
 echo 'false';
}
?>

Interesting. This only gives an error in PHP7.4.0beta1 onward... 3v4l.org/ZvNBA — Nick
– Nick, Commented Sep 26, 2019 at 11:43
I know this is not the answer to your problem, but surely just checking if the element exists is a common enough check that this problem (IMHO) should never matter in real code. — Nigel Ren
– Nigel Ren, Commented Sep 26, 2019 at 11:50
@NigelRen I agree. This is not a problem but sometime we just dont bother checking if every element exist or not depending on the situation. This errors then help so much actually. — Hutarsh
– Hutarsh, Commented Sep 26, 2019 at 12:01
In answer to your now-deleted question, yes, you can make a website in PHP without using classes (object orientation, or OO). OO can be helpful in simplifying complex systems, but you don't have to use it. — halfer
– halfer, Commented Oct 13, 2019 at 10:03

freeek · Accepted Answer · 2019-09-26 11:44:10Z

3

Prior to version 7.4 Undefined index throws for arrays only:

<?php
$info = [];
$value = 'string';
if ($value == $info['firstname']) {
 echo 'true';
} else {
 echo 'false';
}
?>

answered Sep 26, 2019 at 11:44

freeek

9989 silver badges23 bronze badges

Sign up to request clarification or add additional context in comments.

1 Answer 1