根据另一个MySQL表中的信息显示一个表的结果

Question

I've been trying to solve this for a long time now, I give up. I'm getting info from one table, no problem. But I need to exclude some results depending on info from another table.

---

querylist

id | artist         | title 
1  | Kelly Clarkson | Catch My Breath
2  | Nicki Minaj    | Va Va Voom
3  | Jingle         | Jingle100

---

songlist

id | artist         | title           | songtype
1  | Kelly Clarkson | Catch My Breath | S
2  | Nicki Minaj    | Va Va Voom      | S
3  | Jingle         | Jingle100       | I

---

When I get my result from querylist table I want to exclude all songs that do NOT have songtype S from songlist table.

I have this, how do I get songs ONLY with songtype S?

$result = mysql_query("SELECT * FROM queuelist q LEFT JOIN requestlist r ON q.requestid = r.id LEFT JOIN songlist s ON q.songid = s.id ORDER BY q.sortID ASC LIMIT 5 ",$dbcon);

 echo "<table>\n";
 if ($myrow = mysql_fetch_array($result)){
    do{
      $artist = $myrow["artist"]; 
      $title = $myrow["title"];
      echo "<tr><td>$artist - $title</td>";
      echo "</tr>\n";

   }while($myrow = mysql_fetch_array($result));
 }
echo "</table>";

Answer 1

Just to be clear: Adding: WHERE s.songtype='S'

SELECT * FROM queuelist q 
LEFT JOIN requestlist r ON q.requestid = r.id 
LEFT JOIN songlist s ON q.songid = s.id  
WHERE s.songtype='S' 
ORDER BY q.sortID ASC LIMIT 5