CODEWITHSUNDEEP
phparrays
Damien
Damien

Reputation: 33

in_array function is not working

This code should return TRUE value: 

<?php
      $return = in_array(array(1, 2), array(1, 2));
?>

but in_array returns FALSE.

Upvotes: 2

Views: 3188

Answers (10)

dynamic
dynamic

Reputation: 48091

in_array checks if a value exists in an array.

Your $needle doens't exists at all as a value of $haystack

that would be ok if your $haystack was

array(1,2,3,array(1,2))

Notice in this case array(1,2) actually is found inside as expected

If you want to check whenever 2 arrays are equal i suggest you the === operator

($a === $b) // TRUE if $a and $b have the same key/value pairs in the same order and of the same types.

Upvotes: 13

karim79
karim79

Reputation: 342635

Are you interested in intersection?

$arr1 = array(1, 2);
$arr2 = array(1, 2);

$return = array_intersect($arr1, $arr2);

if(count($return) === count($arr1)) {
    // all are present in arr2
}

Upvotes: 3

faken
faken

Reputation: 6852

In your case, the first parameter of in_array should not be an array, but an integer. What you are doing with that code is checking for the presence of an array inside the array, which is not there. A correct form would be:

in_array(1, array(1, 2)); // true

Upvotes: 1

BenWells
BenWells

Reputation: 327

array(1,2) is not in array(1,2) it is array(1,2),

$return = in_array(array(1, 2), array(array(1, 2)));

would return true. (more an extension of yes123's answer)

Upvotes: 2

xkeshav
xkeshav

Reputation: 54016

if second array looks like this

array(array(1, 2));

then return true

Upvotes: 1

Rufinus
Rufinus

Reputation: 30731

you missunderstand in_array see offiziell docs: http://uk.php.net/in_array

<?php
$a = array(array('p', 'h'), array('p', 'r'), 'o');

if (in_array(array('p', 'h'), $a)) {
    echo "'ph' was found\n";
}

if (in_array(array('f', 'i'), $a)) {
    echo "'fi' was found\n";
}

if (in_array('o', $a)) {
    echo "'o' was found\n";
}
?>

Upvotes: 2

Parris Varney
Parris Varney

Reputation: 11478

Your first array isn't contained in the second array, it's equal.

This returns true:

var_dump(in_array(array(1, 2), array(1, 2, array(1, 2))));

Upvotes: 4

Justin Ethier
Justin Ethier

Reputation: 134167

According to the PHP Manual for in_array, the function's syntax is:

bool in_array ( mixed $needle , array $haystack [, bool $strict = FALSE ] )

So you need to supply a $needle value as the first argument. This explains why your example returns FALSE. However, these examples will each return TRUE:

in_array(1, array(1, 2));
in_array(2, array(1, 2));
in_array(array(1, 2), array(1, 2, array(1, 2)))

That said, it might help if you explain exactly what you are trying to do. Perhaps in_array is not the function you need.

Upvotes: 4

Jason McCreary
Jason McCreary

Reputation: 72971

Based on your example, you may want to look into array_intersect(). It compares arrays in a fashion that may better align with your spec.

Upvotes: 5

ecchymose
ecchymose

Reputation: 673

First parameter is the value you're looking for in the second parameter (array) http://php.net/manual/fr/function.in-array.php

Upvotes: 2

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