PHP in_array isn't finding a value that is there

Question

I have an array called $friend_array. When I print_r($friend_array) it looks like this:

Array ( [0] => 3,2,5 )

I also have a variable called $uid that is being pulled from the url.

On the page I'm testing, $uid has a value of 3 so it is in the array.

However, the following is saying that it isn't there:

if(in_array($uid, $friend_array)){
  $is_friend = true;
}else{
  $is_friend = false;

This always returns false. I echo the $uid and it is 3. I print the array and 3 is there.

What am I doing wrong? Any help would be greatly appreciated!

Worth to mention: $is_friend = in_array($uid, $friend_array) is much more readable and does exactly the same. — KingCrunch
– KingCrunch, Commented Aug 6, 2011 at 23:29
The values are coming from SQL. I have a field called "friend_array" and it contains a comma separated list of values. Do I need to do a foreach within my while loop? I am terrible with arrays. — Dex
– Dex, Commented Aug 6, 2011 at 23:33

Andrey Butov · Accepted Answer · 2011-08-06 23:27:27Z

5

Output of

Array ( [0] => 3,2,5 )

... would be produced if the array was created by something like this:

$friend_array = array();
array_push($friend_array, '3,2,5');
print_r($friend_array);

Based on your question, I don't think this is what you meant to do.

If you want to add three values into the first three indexes of the array, do the following:

$friend_array = array();
array_push($friend_array, '3');
array_push($friend_array, '2');
array_push($friend_array, '5');

or, as a shorthand for array_push():

$friend_array = array();
$friend_array[] = '3';
$friend_array[] = '2';
$friend_array[] = '5';

answered Aug 6, 2011 at 23:27

Andrey Butov

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1 Comment

Dex Over a year ago

This is really helpful...I think I need to read more about arrays and see if there's a more efficient way to do this. Thanks for your response!

Peter Ajtai · Accepted Answer · 2011-08-06 23:41:09Z

3

Array ( [0] => 3,2,5 ) means that the array element 0 is a string 3,2,5, so, before you do an is_array check for the $uid so you have to first break that string into an array using , as a separator and then check for$uid:

// $friend_array contains as its first element a string that
// you want to make into the "real" friend array:
$friend_array = explode(',', $friend_array[0]);

if(in_array($uid, $friend_array)){
  $is_friend = true;
}else{
  $is_friend = false;
}

Working example

answered Aug 6, 2011 at 23:41

Peter Ajtai

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Comments

rahlstrom · Accepted Answer · 2011-08-06 23:22:52Z

0

Looks like your $friend_array is setup wrong. Each value of 3, 2, and 5 needs its own key in the array for in_array to work.

Example:

$friend_array[] = 3;
$friend_array[] = 2;
$firned_array[] = 5;

Your above if statement will then work correctly.

answered Aug 6, 2011 at 23:22

rahlstrom

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