根据来自另一个MySQL表的信息显示一个表的结果

Question

我已经试着解决这个问题很长时间了，我放弃了。我从一张桌子上得到信息，没问题。但我需要排除一些结果，这取决于另一个表中的信息。

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

当我从querylist表中得到我的结果时，我想从歌曲列表表中排除所有没有歌曲类型S的歌曲。

我有这个，怎么才能得到只有S型歌曲的歌曲呢？

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

为了清楚起见:添加: WHERE .S. 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