CODEWITHSUNDEEP
phparrays
Matt Elhotiby
Matt Elhotiby

Reputation: 44066

php array_search returning 0 for the first element?

I have this code

$restaurant = array('/restaurant_pos/', '/bar_nightclub_pos/')
$current_page = $_SERVER['REQUEST_URI'];

if (array_search($current_page, $restaurant)) {
    echo "KEEP ME";
}

the problem is the array_search is returning 0 because '/restaurant_pos/' is the first element in the array which is causing the if to fail...any ideas on how to check if the value is in the array without failing on the first element

Upvotes: 18

Views: 14834

Answers (4)

Elangovan Manickasundaram
Elangovan Manickasundaram

Reputation: 21

I have used is_numeric function to achieve the same.

if(is_numeric(array_search($entry, $array))){
            //if true, this block will be executed
 }

Upvotes: 2

Ted
Ted

Reputation: 4166

I think it will be better to use in_array() in this case

$restaurant = array('/restaurant_pos/', '/bar_nightclub_pos/')
$current_page = $_SERVER['REQUEST_URI'];

if (in_array($current_page, $restaurant)) {
    echo "KEEP ME";
}

http://php.net/manual/en/function.in-array.php

Upvotes: 6

vicrec
vicrec

Reputation: 26

From my own experience, if you got, for example:

Array
(
    [1] => Array
        (
            [0] => a
            [1] => b
        )

    [2] => Array
        (
            [0] => b
        )
    [4] => Array
        (   
           [0] => c
        )
)

On array_search("a", $array[3]) !== FALSE) it returns TRUE the same, so to cover all cases, also on null element, it's better use:

if ( (array_search($element, $array) !== FALSE) && ($array) ) {
    echo "found";
}

Hope it helps.

Upvotes: 0

user827080
user827080

Reputation: 

if (array_search($current_page, $restaurant) !== FALSE ) {
    echo "KEEP ME";
}

Manual link: http://php.net/manual/en/function.array-search.php

Upvotes: 44

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