请查看以下查询:
tbl_Contents
Content_Id Content_Title Content_Text
10002 New case Study New case Study
10003 New case Study New case Study
10004 New case Study New case Study
10005 New case Study New case Study
10006 New case Study New case Study
10007 New case Study New case Study
10008 New case Study New case Study
10009 New case Study New case Study
10010 SEO News Title SEO News Text
10011 SEO News Title SEO News Text
10012 Publish Contents SEO News Text
tbl_Media
Media_Id Media_Title Content_Id
1000 New case Study 10012
1001 SEO News Title 10010
1002 SEO News Title 10011
1003 Publish Contents 10012
QUERY
SELECT
C.Content_ID,
C.Content_Title,
M.Media_Id
FROM tbl_Contents C
LEFT JOIN tbl_Media M ON M.Content_Id = C.Content_Id
ORDER BY C.Content_DatePublished ASC
RESULT
10002 New case Study 2014-03-31 13:39:29.280 NULL
10003 New case Study 2014-03-31 14:23:06.727 NULL
10004 New case Study 2014-03-31 14:25:53.143 NULL
10005 New case Study 2014-03-31 14:26:06.993 NULL
10006 New case Study 2014-03-31 14:30:18.153 NULL
10007 New case Study 2014-03-31 14:30:42.513 NULL
10008 New case Study 2014-03-31 14:31:56.830 NULL
10009 New case Study 2014-03-31 14:35:18.040 NULL
10010 SEO News Title 2014-03-31 15:22:15.983 1001
10011 SEO News Title 2014-03-31 15:22:30.333 1002
10012 Publish 2014-03-31 15:25:11.753 1000
10012 Publish 2014-03-31 15:25:11.753 1003
10012将来两次......!
我的查询是从tbl_Contents(连接中的左表)
返回重复的行tbl_Contents中的某些行在tbl_Media中有超过1个关联行。 我需要来自tbl_Contents的所有行,即使tbl_Media中存在Null值但是没有重复记录。
答案 0 :(得分:46)
尝试OUTER APPLY
SELECT
C.Content_ID,
C.Content_Title,
C.Content_DatePublished,
M.Media_Id
FROM
tbl_Contents C
OUTER APPLY
(
SELECT TOP 1 *
FROM tbl_Media M
WHERE M.Content_Id = C.Content_Id
) m
ORDER BY
C.Content_DatePublished ASC
或者,你可以GROUP BY
结果
SELECT
C.Content_ID,
C.Content_Title,
C.Content_DatePublished,
M.Media_Id
FROM
tbl_Contents C
LEFT OUTER JOIN tbl_Media M ON M.Content_Id = C.Content_Id
GROUP BY
C.Content_ID,
C.Content_Title,
C.Content_DatePublished,
M.Media_Id
ORDER BY
C.Content_DatePublished ASC
OUTER APPLY
选择与左表中每一行匹配的单行(或无)。
GROUP BY
执行整个连接,但随后会折叠提供的列上的最终结果行。
答案 1 :(得分:15)
您可以使用group by
的通用SQL执行此操作:
SELECT C.Content_ID, C.Content_Title, MAX(M.Media_Id)
FROM tbl_Contents C LEFT JOIN
tbl_Media M
ON M.Content_Id = C.Content_Id
GROUP BY C.Content_ID, C.Content_Title
ORDER BY MAX(C.Content_DatePublished) ASC;
或者使用相关子查询:
SELECT C.Content_ID, C.Contt_Title,
(SELECT M.Media_Id
FROM tbl_Media M
WHERE M.Content_Id = C.Content_Id
ORDER BY M.MEDIA_ID DESC
LIMIT 1
) as Media_Id
FROM tbl_Contents C
ORDER BY C.Content_DatePublished ASC;
当然,limit 1
的语法因数据库而异。可能是top
。或rownum = 1
。或fetch first 1 rows
。或类似的东西。
答案 2 :(得分:2)
使用DISTINCT标志将删除重复的行。
SELECT DISTINCT
C.Content_ID,
C.Content_Title,
M.Media_Id
FROM tbl_Contents C
LEFT JOIN tbl_Media M ON M.Content_Id = C.Content_Id
ORDER BY C.Content_DatePublished ASC