根據來自另一個MySQL表的信息顯示一個表的結果

Question

我一直試圖解決這個很長一段時間，我放棄了。 我從一張桌子獲得信息，沒問題。 但我需要根據另一個表中的信息排除一些結果。

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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 

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當我從querylist表我的結果，我要排除那些沒有songtype■從表中的SongList所有歌曲。

我有這個，我怎樣才能得到歌曲只與宋型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

只要是明確的：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