Return a MySQL row in a function as an array

Question

I'm trying to create a function to return all of the users in a database as an array. When I print_r the users, I'm getting a multi-dimensional array returned:

Ex:

Array ( [0] => Array ( [0] => Albers [1] => Abbings [2] => Blaine [3]

Code:

   <?php
        session_start();
        function getActiveUsers() {
            $con = mysql_connect("-","-","-");
            if (!$con)
              {
              die('Could not connect: ' . mysql_error());
              }
            else {
                // connected to database successfully
            }
            mysql_select_db("casemanagers", $con);
            $result = mysql_query('select * from users');
            $arr = array();
            while($row= mysql_fetch_assoc($result)){
                $arr[] = $row['Name'];
            }
            return array ($arr);
        }

        $row = array();
        $row = getActiveUsers();
        print_r($row);

    ?>

Can anyone tell me why this is happening or steps I can take to fix it? Thanks.

Gabi Purcaru · Accepted Answer · 2011-08-17 19:43:06Z

2

It's the return line: you should return $arr not array($arr).

What you did there was to create a new array with $arr as its only element.

answered Aug 17, 2011 at 19:43

Gabi Purcaru

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PeeHaa · Accepted Answer · 2011-08-17 19:43:12Z

2

It's because you do:

return array ($arr);

just do:

return $arr;

answered Aug 17, 2011 at 19:43

PeeHaa

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Naftali · Accepted Answer · 2011-08-17 19:42:59Z

1

Try this:

<?php
    session_start();
    function getActiveUsers() {
        $con = mysql_connect("-","-","-");
        if (!$con)
          {
          die('Could not connect: ' . mysql_error());
          }
        else {
            // connected to database successfully
        }
        mysql_select_db("casemanagers", $con);
        $result = mysql_query('select * from users');
        $arr = array();
        while($row= mysql_fetch_array($result)){ //<-- use array instead of assoc
            $arr[] = $row['Name'];
        }
        return $arr; //<-- just return the array, nothing outside it
    }

    $row = array();
    $row = getActiveUsers();
    print_r($row);

?>

answered Aug 17, 2011 at 19:42

Naftali

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