PHP in_array doesnt work right?

Question

So I am trying to see if a set of numbers is in an array.

array(3) { 
     [0]=> array(2) { 
        ["account_ID"]=> string(10) "1781890863" 
        [0]=> string(10) "1781890863" 
      } 
     [1]=> array(2) { 
        ["account_ID"]=> string(10) "2093832999" 
        [0]=> string(10) "2093832999" 
      } 
     [2]=> array(2) { 
        ["account_ID"]=> string(10) "2147483647" 
        [0]=> string(10) "2147483647" } 
      }

I have the array and the values are all there and everything is just dandy. But when I compare using in_array it returns false. I don't quite know why.

class DB(){

    public function getAllID(){
        $result_row = $this->accessDB( 'SELECT account_ID FROM users;');
        return $result_row; 
    }
}

That is the function that I am using to access the database and return the array and then

$app = new DB();
if(isset($_GET['user'])){
    if(in_array($_GET['user'],$app->getAllID())){
        include('account.php');
        echo 'Account DOES exist';
    } else {
        var_dump($app->getAllID());
        echo '<br/>'.$_GET['user'].' does NOT exist.';
    } 
}

Does anyone see why my code here won't work maybe I am just accessing the DB wrong?

return $result_row; will return an associative array so yes the way you are trying will not work — Abhik Chakraborty
– Abhik Chakraborty, Commented Mar 9, 2014 at 19:43
The value of $_GET['user'] should be one of the items of the array. When It is set to one of the items it returns false, when it it set to a number other than that in the array it returns false. — Austin Kregel
– Austin Kregel, Commented Mar 9, 2014 at 19:46

Im0rtality · Accepted Answer · 2014-03-09 19:48:45Z

2

Your method returns array of assoc array, instead of array of int/string (on which your check is working).

Easiest solution - pull account_ID one level up.

public function getAllID(){
    $result_row = $this->accessDB( 'SELECT account_ID FROM users;');
    return array_map(function($entry) { return $entry['account_ID']; }, $result_row); 
}

Better solution - validate with DB query:

// must escape $userId before doing query or you are vulnerable to SQL injection

$this->accessDB("SELECT count(account_ID) FROM users WHERE account_ID = {$userId} LIMIT 1");

answered Mar 9, 2014 at 19:48

Im0rtality

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3 Comments

m79lkm Over a year ago

This is a good solution. Using array_map is much cleaner than the answer I posted. I also like that you mentioned validating the specific user rather than querying for all user ids

Austin Kregel Over a year ago

If I were to validate with a DB query would I still return the array_map or would I return $result_row ?

Im0rtality Over a year ago

As query returns only one field in one row (the count) I would convert that one value to bool. Using var format you gave in question your count value, return (1 == $result_row[0][0]); should work for isUserFound($userId)

AbraCadaver · Accepted Answer · 2014-03-09 19:47:53Z

0

You need to loop through $app->getAllID() because if the $_GET['user'] is there it will be in $app->getAllID()[0], $app->getAllID()[1] etc..

Really, you should do this check for $_GET['user'] in the query with a WHERE clause and return true if it exists and false if not. Why query and return all account_IDs just to check if one exists?

answered Mar 9, 2014 at 19:47

AbraCadaver

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Comments

m79lkm · Accepted Answer · 2014-03-09 19:55:43Z

0

Your function pulling the query results is returning a multi-deminsional array. You can either loop through the array to determine if $_GET['user'] is in each array within or change your function that is pulling the user ids.

In your function loop through your query results and build your array:

$user_ids = array();
foreach ($result_row as $record) {
    $user_ids[] = $record['account_ID'];
}
return $user_ids;

This will create an array of user ids. Then you can use in_array() to determine if the user id exists.

answered Mar 9, 2014 at 19:55

m79lkm

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