PHP In_Array()/Array_Search() Odd Behaviour

PHP in_array() / array_search() odd behaviour

This behaviour of the function in_array() and array_search() is not a bug, but instead well documented behaviour.

Both functions have a 3rd optional parameter called $strict which by default is FALSE:

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

mixed array_search ( mixed $needle , array $haystack [, bool $strict = false ] )

Now what that means is that by default both functions use loosely(==) comparison to compare the values. So they only check if the values are the same after PHP type juggling and without checking the type. Because of that in your example TRUE == "any none emtpy string" evaluates to TRUE.

So by setting the 3rd parameter to TRUE while calling the function you say that PHP should use strict(===) comparison and it should check value AND type of the values while comparing.

See this as a reference: How do the PHP equality (== double equals) and identity (=== triple equals) comparison operators differ?

Weird in_array() behaviour

Use third parameter of in_array to force strict match

<?php

$_aOperatorsOneOptin = array('DE-010', 'DE-005');

$bMatchPaymentOperator = in_array(0 , $_aOperatorsOneOptin, true);

if($bMatchPaymentOperator == true)
    echo 'found';

Weird in_array() behaviour

Use third parameter of in_array to force strict match

<?php

$_aOperatorsOneOptin = array('DE-010', 'DE-005');

$bMatchPaymentOperator = in_array(0 , $_aOperatorsOneOptin, true);

if($bMatchPaymentOperator == true)
    echo 'found';

Weird in_array() behaviour

Use third parameter of in_array to force strict match

<?php

$_aOperatorsOneOptin = array('DE-010', 'DE-005');

$bMatchPaymentOperator = in_array(0 , $_aOperatorsOneOptin, true);

if($bMatchPaymentOperator == true)
    echo 'found';

Weird array_search behaviour

This is because array_search uses == to compare things. This makes PHP convert the operands so that their types match.

1 == "I want to find this text"
"aa" == "I want to find this text"
0 == "I want to find this text"

In the 1st and 3rd ones, PHP needs to convert "I want to find this text" to a number so it can compare. When converting a string to a number, PHP reads from the beginning of the string and stops at the first non-numeric character. So "I want to find this text" is converted to 0.

So the comparisons made are

1 == "I want to find this text" => 1 == 0 => false
"aa" == "I want to find this text" => false
0 == "I want to find this text" => 0 == 0 => true

And, that's why you get 2.

To fix this, do this: array_search("I want to find this text", $ray, true)

The 3rd parameter tells array_search to use === instead. This does not convert types, instead it compares them too. That will give you FALSE since nothing matches "I want to find this text" in both type and value.

Unexpected in_array() behaviour

The function in_array() by default performs a loose comparison and this causes PHP to evaluate the expression (0 == "article") as TRUE.

You need to use the strict parameter for in_array() to also check the types of the needle in the haystack:

var_dump( in_array('article', $enabled, true) );

Demo!

in_array() returns wrong values?

in_array you have to set type

Syntax:

in_array(search,array,type);

var_dump(in_array(0, array('a', 'ab', 'abc'), true));

Note: If the search parameter is a string and the type parameter is set to TRUE, the search is case-sensitive.

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

in_array returns True from a value that not exists

in_array() check for the values of the arrays.

If you are setting a value as true, it will return true because of that, unless, as mentioned by @AymDev before, you set the third parameter to be strict.

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