我有两个表,分别包含员工数据TableA
和TableB
,我基于2个id(其中一个userID
和另一个mID
(基于月份的ID))使用LEFT OUTER JOIN
到大约20%的结果中返回NULL,因为TableB
不完整。我想要-如果可能的话,一个查询来检测联接是否找不到匹配项,并将一个月减去到mID
中,以便联接可以用旧数据覆盖至少一部分丢失数据。
我不知道这是否是一种过于复杂的查询,但是我想到了类似的东西:
SELECT T1.*, T2.*
FROM TABLEA
LEFT OUTER JOIN TABLEB
ON T2.USERID = T1.USERID AND (CASE WHEN (T2.HID = T1.HID) = NULL THEN (T2.HID = T1.HID-1))
感谢任何帮助。
答案 0 :(得分:0)
我认为这是您真正想要做的;缺点是逻辑更多,每列都可以重复,但是可以提供很好的控制。
DECLARE @TableA TABLE (USERID INT,HID INT,SomeData VARCHAR(20))
DECLARE @TableB TABLE (USERID INT,HID INT,SomeData VARCHAR(20))
INSERT INTO @TableA(USERID,HID,SomeData) SELECT 1,5,'Now'
INSERT INTO @TableA(USERID,HID,SomeData) SELECT 2,5,NULL
INSERT INTO @TableA(USERID,HID,SomeData) SELECT 3,5,NULL
INSERT INTO @Tableb(USERID,HID,SomeData) SELECT 2,4,'Now-1'
INSERT INTO @Tableb(USERID,HID,SomeData) SELECT 2,3,'Now-2'
INSERT INTO @Tableb(USERID,HID,SomeData) SELECT 3,4,'Now-1'
SELECT
t1.USERID, T1.Hid AS [Current HID]
,
CASE
WHEN T1.SomeData IS NOT NULL THEN T1.SomeData
WHEN T2.USERID IS NOT NULL THEN T2.SomeData
WHEN T3.USERID IS NOT NULL THEN T3.SomeData
ELSE T1.SomeData
END AS [Most Recent SomeData]
FROM @TABLEA T1
LEFT JOIN @TABLEB T2 ON T2.USERID = T1.USERID AND T2.HID = T1.HID
LEFT JOIN @TABLEB T3 ON T3.USERID = T1.USERID AND T3.HID = T1.HID-1
答案 1 :(得分:0)
通过使用case语句,您处在正确的轨道上,但这有点不合时宜。警告!我没有进行测试,但我相信过去也遇到过类似的情况。
SELECT T1.*
,T2.*
FROM TABLEA t1
LEFT JOIN TABLEB t2 ON T2.USERID = T1.USERID
AND T2.HID = CASE
WHEN T2.HID = T1.HID then t1.hid
else T1.HID - 1
end