PHP Array_search won't find an existing value

Question

I'm trying to use array_search on a simple array :

        echo "<br> Search for : -".ucfirst(strtolower(trim($rowData[0][2]))).'-';

here is the index of the value ( I know it but I want PHP to find it for me >< ):

        echo '<br>-'.$listMetiers[0].'-';

Here is the full array :

        echo '<pre>';
        print_r($listMetiers);
        echo '</pre>';

        $id_metier = array_search(ucfirst(strtolower(trim($rowData[0][2]))),$listMetiers);

        if(!$id_metier)
        {
          echo ' NOT FOUND !<br>';
          $id_metier = -666;
        }
        else
        {
          echo 'GOOD : '.$id_metier.'<br>';
        }

The value is in the array but array_search don't find it ! Look what I have when executing this code :

What is going on ?

Your found index is 0 and !0 is == TRUE, so it enters the if statement; Check if $id_metier is equals FALSE or not — Rizier123
– Rizier123, Commented Oct 17, 2015 at 15:04

Nenad Mitic · Accepted Answer · 2015-10-17 16:22:04Z

3

array_searh is returning 0, and PHP is treating that as a falsy value.

You should change this

!$id_metier

into this

$id_metier === false

edited Oct 17, 2015 at 16:22

answered Oct 17, 2015 at 15:08

Nenad Mitic

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2 Comments

Federkun Over a year ago

correct, but the op wants to check if it doesn't exist, so === false

Seba99 Over a year ago

Yup @Rizier123 commented on this ! :) That's it but better change into $id_metier === false ;)

Ghilas BELHADJ · Accepted Answer · 2015-10-17 15:06:14Z

1

Your test is wong, use:

if($id_metier < 0){ // not found
// ...
} else {            // found
// ...
}

answered Oct 17, 2015 at 15:06

Ghilas BELHADJ

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