PHP first index gets treated as string

Question

I will keep this brief, this is my code:

<?php

$arr = array(
    1,
    2,
    3,
    "x" => "aaa", 
    "y" => "bbb"
);

foreach($arr as $k => $v){

    if($k !="y" && $k != "x") {
        echo $v . " ";
    }

}

?>

and this is the result:

2 3

and it acts as though this:

$example = 0;
if ($example == 'x' || $example == 'y') {
    echo "true";
}

would echo out "true" instead of nothing.

Basically, my question is: why does it skip echo-ing out the first element of the array, if 0 does not equal "x" or "y"?

Reading it back, the title of the question seems very misleading, but i'm not sure what to title it. — aakk
– aakk, Commented Oct 17, 2017 at 18:54
It's an issue with != being more lenient than !== because of the 0 value of the key. Try the latter, and you'll get what you're looking for. — aynber
– aynber, Commented Oct 17, 2017 at 18:55
@aynber Yes, I think it is, but I wasn't sure where the problem was before asking the question, now that I know, I would agree with you — aakk
– aakk, Commented Oct 17, 2017 at 19:00

Unamata Sanatarai · Accepted Answer · 2017-10-17 19:00:23Z

4

The == sorts out the types for you.

0 is an int, so it is going to cast y and x to an int. Which is not parseable as one, and will become 0. A string '0x' would become 0, and would match!

Go with !==

http://sandbox.onlinephpfunctions.com/code/bed936181e386ddfe75e4ef92771bf61ad2e7915

<?php

$arr = array(
    1,
    2,
    3,
    "x" => "aaa", 
    "y" => "bbb"
);

foreach($arr as $k => $v){

    if($k !=="y" && $k !== "x") {
        echo $v . " ";
    }

}
// result:
// 1 2 3

source

edited Oct 17, 2017 at 19:00

answered Oct 17, 2017 at 18:55

Unamata Sanatarai

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2 Comments

aynber Over a year ago

0 is also a false-y value. I think it's using that more than using casting the letter to an int.

Ingo Over a year ago

I wouldn't have answered had I seen this answer, thank you @Unamata Sanatarai. Great answer!

Ingo · Accepted Answer · 2017-10-17 19:32:25Z

2

That is a good one, I think I found the fix, instead of using != in both of your evaluations in the if statement, try using !== so, change your if statement to:

if($k !== "y" && $k !== "x")

See Unamata Sanatarai answer above, explained very well.

edited Oct 17, 2017 at 19:32

answered Oct 17, 2017 at 19:26

Ingo

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