php array_search 0 index

Question

It seems that that you can not use the search_array function in PHP to search the 0 index AND have it evalute as true.

Consider this code for example:

$test=array(100, 101, 102, 103);

if($key=array_search(100,$test)){

     echo $key;

}

else{

     echo "Not found";

}

The needle '100' is found in the haystack and the key is returned as 0. So far so good, but then when I evaluate whether the search was successful or not it fails because the returned value is 0, equal to false!

The php manual suggests using '!==' but by doing so the key (array index) is not returned, instead either 1 or 0 is returned:

if($key=(array_search(103,$test)!== false)){

}

So how can I successfully search the array, find a match in the 0 index and have it evaluate as true?

Did you not see the BIG RED BOX? Try to read the documentation and understand the language before you ask such questions. — cmbuckley
– cmbuckley, Commented Apr 10, 2013 at 19:16
This is just bad and lazy code. Why do you need to test and get in the same if ? — Sergio Abreu
– Sergio Abreu, Commented Aug 30, 2024 at 11:27

user229044 · Accepted Answer · 2018-01-04 12:48:02Z

46

This is explicitly mentioned in the docs. You need to use === or !==:

$key = array_search(...);

if ($key !== false) ...

Otherwise, when $key is 0, which evaluates to false when tested as a boolean.

edited Jan 4, 2018 at 12:48

answered Apr 10, 2013 at 19:13

user229044♦

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5 Comments

Niet the Dark Absol Over a year ago

You could of course use if( ($key = array_search(...)) !== false)

user2056238 Over a year ago

I read several threads on this, read the manual over and over but couldn't find the answer. I think I had the '!==' in the wrong part of my code. Now working - thanks!

Ashish Rawat Over a year ago

but in this manner if I use code like if(array_search(...) === true), then why it is not working. (=== true is also string comparison) can anybody explain me

user229044 Over a year ago

@EstailSe Your code makes no sense. array_search can never return boolean true, so no, it will not work, and it's not clear what you're trying to accomplish.

Ashish Rawat Over a year ago

okay I got your point! thanks, I was comparing index number with Boolean, there I was wrong!

Sammitch · Accepted Answer · 2017-12-06 19:58:30Z

5

The conditional in your second example block gives execution order priority to the !== operator, you want to do the opposite though.

if (($key = array_search(100,$test)) !== false) {

!== has higher precedence than == which makes the parentheses necessary.

edited Dec 6, 2017 at 19:58

Sammitch

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answered Apr 10, 2013 at 19:15

Explosion Pills

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1 Comment

Sammitch Over a year ago

Note: This is the most correct in here. If I could transfer the upvotes from my awful, old, naive, now-deleted answer I would.

Andrej Bestuzhev · Accepted Answer · 2013-04-10 19:17:43Z

1

$key = array_search($what, $array);
if($key !== false and $array[$key] == $what) {
 return true;
}

it's more secure

answered Apr 10, 2013 at 19:17

Andrej Bestuzhev

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Comments

Peter Kiss · Accepted Answer · 2013-04-10 19:15:18Z

0

if(($key = array_search(103,$test)) !== false){

}

answered Apr 10, 2013 at 19:15

Peter Kiss

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Comments

rorra · Accepted Answer · 2013-04-11 00:16:26Z

0

$test=array(100, 101, 102, 103);

if (($key = array_search(100,$test)) === false) {
    echo "Not found";
} else{
    echo $key;
}

answered Apr 11, 2013 at 0:16

rorra

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