CODEWITHSUNDEEP
phparraysloopsforeach
datether
datether

Reputation: 163

PHP iterating through array returned from a function

Hi I have a class file which contains a function to retrieve users from a mySQL database. The code is as following :

class userdbase
{
    public function fetchUsers()
    {
        $db = new mysqli('localhost', 'user', 'pass', 'dbase');

        if($db->connect_errno > 0)
        {
            die('Unable to connect to database [' . $db->connect_error . ']');
        }

        $sql = "SELECT username from `users`";

       if(!$result = $db->query($sql))
       {
           die('There was an error running the query [' . $db->error . ']');
       }

       while($row = $result->fetch_assoc())
       {
           $users[] = $row['username'];
       }
       $result->free();
       $db->close();
       return array($users);
    }
}

The code above returns an array as following using print_r($dbase->fetchUsers()) : 

Array
(
[0] => Array
    (
        [0] => user1
        [1] => user2
        [2] => user3

        )

)

I am having trouble looping through the array from the main file where i reference to the function in the class . The code i am using is as following :

<?php

require("dbase.php");

$dbase = new userdbase();

 foreach($dbase->fetchUsers() as $result)  //this doesn't work
 {
    print "$result"; 
 }
?>

Can someone more versed in PHP point me out the correct way to do this ? Thanks in advance !

Upvotes: 2

Views: 68

Answers (2)

Narendrasingh Sisodia
Narendrasingh Sisodia

Reputation: 21437

Instead of 

return array($users);

only return

return $users;

and simply apply your foreach loop

Upvotes: 1

Nicholas Alexander
Nicholas Alexander

Reputation: 359

When you build the array of users[] here:

while($row = $result->fetch_assoc())
{
    $users[] = $row['username'];
}

$users is an array, but when you return it, wrapping it in array() means you are returning an array of arrays which is not what you then process.

Solution is just return $users.

Upvotes: 1

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