PHP In_array function Use instructions and in_array need to note the local instructions

In_array
(PHP 4, PHP 5)

in_array-checks if a value exists in the array

Description

BOOL In_array (mixed $needle, array $haystack [, BOOL $strict])

Searches for needle in haystack and returns TRUE if found, otherwise FALSE.

If the value of the third parameter strict is TRUE then the In_array () function also checks whether the needle type is the same as in haystack.

Note: If needle is a string, the comparison is case-sensitive.

Note: Needle is not allowed to be an array until PHP version 4.2.0.

Example #1 In_array () example

<?php $os = Array ("Mac", "NT", "Irix", "Linux"); if (In_array ("Irix", $os)) {echo "Got Irix";} if (In_array ("Mac", $os)) {echo "Got mac";}?>

The second condition fails because In_array () is case-sensitive, so the above program appears as:
Got Irix

Example #2 In_array () strict type checking example

<?php $a = array (' 1.10 ', 12.4, 1.13); if (In_array (' 12.4 ', $a, true)) {echo "' 12.4 ' found with strict check\n";} if (In_array (1.13, $a, true)) {echo "1.13 fo und with strict check\n "; }?>

The example above will output:

1.13 Found with strict check

Example #3 In_array () using Arrays as needle

<?php $a = array (' P ', ' h '), Array (' P ', ' r '), ' o '); if (In_array (' P ', ' h '), $a) {echo "' ph ' was found\n";} if (In_array (' f ', ' I '), $a) {echo "' fi ' is found\ n "; } if (In_array (' o ', $a)) {echo "' O ' was found\n";}?>

The example above will output:

' ph ' was found
' O ' was found

Places to be aware of:

If:

First declare an array as:

$arr = Array (*);

Then there are:

In_array (0, $arr) = = True

Puzzling! {Weak language}

Workaround:
In_array (strval (0), $arr, True))