MySQL query WHERE statement which is a query itself?

So I'll try to clearly explain my goal first:

First, I want to query one table in my database for a list of usernames.

Second, I want to take those usernames, and query another table in the database and return only the rows in which these usernames appear in the username field.

Finally, I want to take this output (in JSON array form right now), and return it to the requesting client.

My query looks like this right now:

$query = mysql_query("SELECT * FROM tagusers WHERE username = 
    (SELECT userA FROM friendtable WHERE userB = '$username')");

This works when the WHERE statement yields 1 result. So if I only get one returned userA, it works fine. But if I get multiple, I get a bunch of errors.

Here is the code in its entirety:

if (isset($_POST['username'])) {
$username = $_POST['username'];


$connect = mysql_connect("localhost", "root", "");
mysql_select_db("TagDB");

$query = mysql_query("SELECT * FROM tagusers WHERE username = 
    (SELECT userA FROM friendtable WHERE userB = '$username')");

    }

while ($e = mysql_fetch_assoc($query)) {
    $output[] = $e;
}

$output = json_encode($output);

print $output;

I get the following error on the query line:

*Warning: mysql_query() [function.mysql-query]: Unable to save result set in C:\w开发者_开发问答amp\www\tag\appgetfriendinfo.php on line 21*

So all I really need to know is, how would I write that query in MySQL so that I get returned an array of rows?

You don't need a subquery at all, you'll usually get better performance out of a join. Make sure you have indexes defined on tagusers.username, friendtable.userA and friendtable.userB

SELECT
  tagusers.*
FROM
  tagusers
INNER JOIN
  friendtable  
ON
  tagusers.username = friendtable.userA
AND
  friendtable.userB = '$username'

Use the IN keyword.

$query = mysql_query("SELECT * FROM tagusers WHERE username IN 
    (SELECT userA FROM friendtable WHERE userB = '$username')");

Use either the IN clause, or a JOIN like in this example:

$query = sprintf("SELECT tu.* 
                    FROM TAGUSERS tu 
                    JOIN FRIENDTABLE ft ON ft.usera = tu.username
                   WHERE ft.userB = '%s'",
                  mysql_real_escape_string($_POST['username']));

$result = mysql_query($query);

$output = json_encode($result);