PHP - in_array isn't finding my value

Question

I'm a newb and I got a problem with in_array...

So this is my array $allUsers (received by a SQL-Query for Usernames)

    Array
(
    [0] => Array
        (
            [name] => test
        )

    [1] => Array
        (
            [name] => test2
        )

    [2] => Array
        (
            [name] => admin
        )

    [3] => Array
        (
            [name] => kingChräbi
        )

Now If a new member wants to register, I want to check in this array if it's already existent:

if(!in_array($username,$allUsers)){....

eventhough it is to when $username is NOT in $allUsers do .... it's just skipping to else also if the user is existing :(

$username is set before with

$username = $_POST['name'];

and working as it should (i can echo it without a problem, is exactly test or test2 without whitespace or anything)

I really looked around alot, but I can't find anything like my problem here... Could you please help me?

Thanks

`in_array()` requires a **value** and **a single-dimensional array**. You're currently passing a 2 d array. That's why it doesn't work. — Amal Murali, Mar 24 '14 at 16:29
alright, So I should write a function returning true / false with a foreach in it. — Michael, Mar 24 '14 at 16:32

Your Common Sense · Accepted Answer · 2014-03-24T16:36:32.713

2

although question itself is quite silly, as you have to realize what array you are working with, the quick solution, based on PDO tag, would be as follows: instead of fetchAll() use fetchAll(PDO::FETCH_COLUMN)

Or, rather you need to learn SQL as well, and find users not by means of selecting them ALL from database which makes no sense, but by asking a database to find a user for you

$stm = $pdo->prepare("SELECT * FROM table WHERE name=?");
$stm->execute(array($_POST['name']));
$user = $stm->fetch();

if ($user) { // <---HERE YOU GO

edited Mar 24 '14 at 16:36

answered Mar 24 '14 at 16:31

Your Common Sense

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Got that! Thank you, sir! – Michael Mar 24 '14 at 16:39
Hope you got that you have to use second option only. – Your Common Sense Mar 24 '14 at 16:40

score 0 · Answer 2 · answered Mar 24 '14 at 16:34

The in_array() does not work with multi-dimensional arrays. You better flatten your array and then do a search for the keyword.

$it = new RecursiveIteratorIterator(new RecursiveArrayIterator($arr)); //<-- Pass your array here
$new_arr = array();
foreach($it as $v) {
    $new_arr[]=$v;
}
if(in_array('test',$new_arr))
{
    echo "Exists !";
}

Working Demo

score 0 · Answer 3 · answered Mar 24 '14 at 16:39

You are searching a 2-D array for a value under the key of "name".
Using array_map() or simple foreach loop should work -

$username = "admin";
$key = "name";
if(!in_array($username,array_map(function($v)use($key){return $v[$key];},$allUsers))){
    echo "No found";
}else{
    echo "Found";
}

score -1 · Answer 4 · answered Mar 24 '14 at 16:36

-1

If you are using:

While ($ row=mysql_fetch_assoc ($ result) {

$ data [] = $ row }

Remove the [] to not create a multidimensional array.

answered Mar 24 '14 at 16:36

Ryguydg

93
1
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PHP - in_array isn't finding my value

4 Answers4