2010-12-06 79 views
1

我想了解我的課程在sql中的別名,但我不完全理解命令。作爲工作的一部分,我必須這樣做:幫助別名sql

  1. 瞭解複雜的SQL查詢 (a)選擇所有列/從主道和錄音師基礎上,音響工程師ID連接兩個和使用別名爲每場表。

我得在這裏:

SELECT * 
FROM SOUNDENGINEER AS s 
INNER JOIN MASTERTRACK AS m ON m.EDITED_BY,s.SOUND_ENG_ID; 

,但現在我堅持請幫助

回答

1
SELECT * FROM SOUNDENGINEER s INNER JOIN MASTERTRACK m ON m.EDITED_BY = s.SOUND_ENG_ID; 

這似乎工作

5

這是正確的,但對條款應該是這樣的在m.EDITED_BY = s.SOUND_ENG_ID;

SELECT * FROM SOUNDENGINEER AS s 
INNER JOIN MASTERTRACK AS m ON m.EDITED_BY = s.SOUND_ENG_ID; 
+0

當我輸入這個我得到這個錯誤:SQL> SELECT * FROM SOUNDENGINEER爲S INNER JOIN MASTERTRACK爲M ON m.EDITED_BY = s.SOUND_ENG_ID; SELECT * FROM SOUNDENGINEER AS s * 錯誤在第1行: ORA-00933:SQL命令未正確結束 – learner123 2010-12-06 14:32:55

+0

在查詢結尾是否包含*號? 「SELECT * FROM SOUNDENGINEER as s INNER JOIN MASTERTRACK as m ON m.EDITED_BY = s.SOUND_ENG_ID; SELECT * FROM SOUNDENGINEER AS s」is pretty legal dude dude,so remove the last * from you query if the case。 – 2010-12-06 14:53:26

+0

最後沒有星星只是顯示錯誤 – learner123 2010-12-06 15:03:28

2
SELECT * FROM SOUNDENGINEER AS s INNER JOIN MASTERTRACK AS m ON m.EDITED_BY = s.SOUND_ENG_ID; 
1

差不多吧。嘗試:

SELECT * 
FROM SOUNDENGINEER AS s 
INNER JOIN MASTERTRACK AS m ON m.EDITED_BY = s.SOUND_ENG_ID; 

您使用逗號可能會建議你試圖替代等連接的語法,這將是:

SELECT * 
FROM SOUNDENGINEER AS s, MASTERTRACK AS m 
WHERE m.EDITED_BY = s.SOUND_ENG_ID; 

我更喜歡前者,不過,因爲你明確地說明你以INNER JOIN聲明加入表格。