Question

我一直试图解决这个问题很长一段时间，我放弃了。我从一张桌子上获取信息，没问题。但我需要根据另一个表中的信息排除一些结果。

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表中获取结果时，我想从歌曲列表中排除所有没有歌曲类型S的歌曲。

我有这个，我如何才能获得歌曲S的歌曲？

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

要明确：添加：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

根据另一个MySQL表中的信息显示一个表的结果

1 个答案: