Unable to set variable type in PHP

Question

Let's say I have an array like this:

$array = array ('1','2','3','4');

And I want to add parts of given string (even it consists of numbers) to the array if it isn't already exist in array. For example, if the given string is 1234, it should add 12 , 23 and 34. I wrote simple code for it like this:

$string = '1234';
$string_length = 4;
$array = array ('1','2','3','4');
for($i=0;$i<$string_length-1;$i++){
    $element = substr($string,$i,2);
    if (!in_array($element, $array)){
        array_push($array, $element);
    }
}

It works well when I try 1234 as $string. But when I try 1024 as $string, it returns the array like this:

Array
(
    [0] => 1
    [1] => 2
    [2] => 3
    [3] => 4
    [4] => 10
    [5] => 24
)

As you see, after [4] => 10, there must be [5] => 02, but there isn't it.

The cause is 02 is equal to 2 (which is in first given array) as a integer. That's why the code skipped it. But I need to add it.

For solving problem I set the type of $element as a string. So I added settype($element,'string'); after $string_length in my code, but nothing changed.

What can I do ?

Answer 1

try:

if (!in_array($element, $array, true)){

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

Answer 2

Use the following code:
Notice I am passing true to the in_array function. It does a strict comparison. http://www.php.net/manual/en/function.in-array.php

$string = '1024';
$string_length = 4;
$array = array ('1','2','3','4');
for($i=0;$i<$string_length-1;$i++){
    $element = substr($string,$i,2);
    if (!in_array($element, $array, true)){
        array_push($array, $element);
    }
}