Sql 选择列上的最大值,然后选择依赖于第一个值的列中的最大值
我有这样的桌子:Sql 选择列上的最大值,然后选择依赖于第一个值的列中的最大值,sql,sql-server,sql-server-2005,max,Sql,Sql Server,Sql Server 2005,Max,我有这样的桌子: CREATE TABLE #Test ( ParentID int, DateCreated DATETIME, ItemNo int ) INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-10-01 00:00:00.000',0) INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-10-
CREATE TABLE #Test
(
ParentID int,
DateCreated DATETIME,
ItemNo int
)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-10-01 00:00:00.000',0)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-10-01 00:00:00.000',1)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-05-01 00:00:00.000',2)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-05-01 00:00:00.000',3)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (2,'2008-06-01 00:00:00.000',3)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (2,'2008-06-01 00:00:00.000',4)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (2,'2008-04-01 00:00:00.000',6)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (2,'2008-04-01 00:00:00.000',8)
SELECT *
FROM #Test t
JOIN
(
If I could get maximum row here somehow that would be great
) maxt
ON t.ParentID = maxt.ParentID
JOIN SomeOtherTable sot
ON sot.DateCreated = maxt.MaxDateCreated
AND sot.ItemNo = maxt.MaxItemNo
GROUP BY
sot.Something
我需要一种方法来选择在同一parentID上使用最高ItemNo创建的最高日期,如果可以在查询中使用如下解决方案:
CREATE TABLE #Test
(
ParentID int,
DateCreated DATETIME,
ItemNo int
)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-10-01 00:00:00.000',0)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-10-01 00:00:00.000',1)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-05-01 00:00:00.000',2)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (1,'2008-05-01 00:00:00.000',3)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (2,'2008-06-01 00:00:00.000',3)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (2,'2008-06-01 00:00:00.000',4)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (2,'2008-04-01 00:00:00.000',6)
INSERT INTO #Test(ParentID, DateCreated, ItemNo) VALUES (2,'2008-04-01 00:00:00.000',8)
SELECT *
FROM #Test t
JOIN
(
If I could get maximum row here somehow that would be great
) maxt
ON t.ParentID = maxt.ParentID
JOIN SomeOtherTable sot
ON sot.DateCreated = maxt.MaxDateCreated
AND sot.ItemNo = maxt.MaxItemNo
GROUP BY
sot.Something
为了澄清结果应该是什么样的:
ParentID DateCreated ItemNo ParentID MaxDateCreated MaxItemNo
1, '2008-10-01 00:00:00.000' ,0 1, '2008-10-01 00:00:00.000',1
1, '2008-10-01 00:00:00.000' ,1 1, '2008-10-01 00:00:00.000',1
1, '2008-05-01 00:00:00.000' ,2 1, '2008-10-01 00:00:00.000',1
1, '2008-05-01 00:00:00.000' ,3 1, '2008-10-01 00:00:00.000',1
2, '2008-06-01 00:00:00.000' ,3 2, '2008-06-01 00:00:00.000',4
2, '2008-06-01 00:00:00.000' ,4 2, '2008-06-01 00:00:00.000',4
2, '2008-04-01 00:00:00.000' ,6 2, '2008-06-01 00:00:00.000',4
2, '2008-04-01 00:00:00.000' ,8 2, '2008-06-01 00:00:00.000',4
如果需要此DateCreated的最大日期和最大ItemNo:
select ParentId,
DateCreated as MaxDateCreated,
ItemNo as MaxItemNo
from
(select PArentID,DateCreated,ItemNo,
Row_Number() OVER (PARTITION BY ParentID
ORDER BY DateCreated DESC,
ItemNo Desc) as RN
from #Test
) t3
where RN=1
UPD
为了得到问题中提到的结果,你应该加入#测试,比如:
SELECT *
FROM Test t
JOIN
(
select ParentId,
DateCreated as MaxDateCreated,
ItemNo as MaxItemNo
from
(select PArentID,DateCreated,ItemNo,
Row_Number() OVER (PARTITION BY ParentID
ORDER BY DateCreated DESC,
ItemNo Desc) as RN
from test
) t3
where RN=1
) maxt
ON t.ParentID = maxt.ParentID
什么是MaxItemNo,为什么不是3和8而不是1和4?@t-clausen.dk MaxItemNo是最大ItemNo,如果日期在同一个参数上是最大的,则父1需要的组合是“2008-10-01 00:00.000”,1,因为这是ParentID的最大创建日期中的最大ItemNo1@t-clausen.dk因为我需要DateCreated和ItemNo的最高组合,因为DateCreated具有第一优先级,所以“2008-10-01 00:00:00.000”,1>“2008-05-01 00:00:00.000”,3我只看到表中的两行result@t-clausen.dk要获得所需的结果,您应该使用#Test on ParentId连接此表。我已经更新了我的答案。