title: Joining two tables without returning unwanted row
body: <p>My table structure looks like this:</p>
  
  <pre><code>      tbl.users                       tbl.issues
  +--------+-----------+   +---------+------------+-----------+
  | userid | real_name |   | issueid | assignedid | creatorid |
  +--------+-----------+   +---------+------------+-----------+
  |   1    |   test_1  |   |    1    |     1      |     1     |
  |   2    |   test_2  |   |    2    |     1      |     2     |
  +--------+-----------+   +---------+------------+-----------+
  </code></pre>
  
  <p>Basically I want to write a query that will end in a results table looking like this:</p>
  
  <pre><code>                           (results table)
  +---------+------------+---------------+-----------+--------------+
  | issueid | assignedid | assigned_name | creatorid | creator_name |
  +---------+------------+---------------+-----------+--------------+
  |    1    |     1      |    test_1     |    1      |    test_1    |
  |    2    |     1      |    test_1     |    2      |    test_2    |
  +---------+------------+---------------+-----------+--------------+
  </code></pre>
  
  <p>My SQL looks like this at the moment:</p>
  
  <pre><code>SELECT 
    `issues`.`issueid`,
    `issues`.`creatorid`,
    `issues`.`assignedid`,
    `users`.`real_name`
  FROM `issues`
  JOIN `users` 
    ON ( `users`.`userid` = `issues`.`creatorid` )
    OR (`users`.`userid` = `issues`.`assignedid`)
  ORDER BY `issueid` ASC
  LIMIT 0 , 30 
  </code></pre>
  
  <p>This returns something like this:</p>
  
  <pre><code>                (results table)
  +---------+------------+-----------+-----------+
  | issueid | assignedid | creatorid | real_name |
  +---------+------------+-----------+-----------+
  |    1    |     1      |     1     |   test_1  |
  |    2    |     1      |     2     |   test_1  |
  |    2    |     1      |     2     |   test_2  |
  +---------+------------+-----------+-----------+
  </code></pre>
  
  <p>Can anyone help me get to the desired results table?</p>