我有一张像
这样的表格id f1
--------------
1 2000-01-01
1 2001-01-01
1 2002-01-01
1 2003-01-01
我想在一行中说出最新的3个日期
CREATE TABLE Test
(
id INT NOT NULL,
f1 DATETIME NOT NULL,
)
INSERT INTO Test (id, f1) VALUES (1, '1/1/2000')
INSERT INTO Test (id, f1) VALUES (1, '1/1/2001')
INSERT INTO Test (id, f1) VALUES (1, '1/1/2002')
INSERT INTO Test (id, f1) VALUES (1, '1/1/2003')
SELECT T1.* FROM Test as T1
正在尝试类似
的内容 SELECT T1.*,T2.*
FROM Test AS T1
LEFT OUTER JOIN Test AS T2 ON T1.id = T2.id AND (T2.f1 > T1.f1)
答案 0 :(得分：4)
虽然我不确定如何将它们分成一行，但您可以从：
开始SELECT * FROM Test ORDER BY f1 DESC LIMIT 3
这应该会给你一个结果：
id f1
1 2003-01-01
1 2002-01-01
1 2001-01-01
但是，将它们放在一排可能会有点困难......
答案 1 :(得分：3)
在sql server中你可以select top 3 * from Test order by f1 desc。其他DBMS具有类似的可能性，例如MySql的limit，Oracle的rownum等。
答案 2 :(得分：2)
您可以使用ORDER BY，TOP和PIVOT的组合执行此操作，至少在SQL Server上是这样。似乎许多其他答案都忽略了结果“需要一行”的必要性。
答案 3 :(得分：2)
（即使它们都是相同的值，这也会得到前三个日期）
with TestWithRowNums(f1, row_num) as
(
select f1, row_number() over(order by [f1] desc) as row_num from test
)
select
(select [f1] from TestWithRowNums where row_num = 1) as [Day 1],
(select [f1] from TestWithRowNums where row_num = 2) as [Day 2],
(select [f1] from TestWithRowNums where row_num = 3) as [Day 3]
这将为您提供前三个DISTINCT日期
with TestWithRankNums(f1, rank_num) as
(
select f1, dense_rank() over(order by [f1] desc) as rank_num from test
)
select
(select top 1 [f1] from TestWithRankNums where rank_num = 1) as [Day 1],
(select top 1 [f1] from TestWithRankNums where rank_num = 2) as [Day 2],
(select top 1 [f1] from TestWithRankNums where rank_num = 3) as [Day 3]
在SQL Server 2005中尝试此操作
--to get top three values even if they are the same
select [1] as Day1, [2] as Day2, [3] as Day3 from
(select top 3 f1, row_number() over(order by [f1] desc) as row_num from test) src
pivot
(
max(f1) for row_num in([1], [2], [3])
) as pvt
--to get top three distinct values
select [1] as Day1, [2] as Day2, [3] as Day3 from
(select f1, dense_rank() over(order by [f1] desc) as row_num from test) src
pivot
(
max(f1) for row_num in([1], [2], [3])
) as pvt
答案 4 :(得分：1)
您要做的事情称为数据透视表，这是一篇关于如何操作的文章：
http://asktom.oracle.com/pls/asktom/f?p=100:11:0::NO::P11_QUESTION_ID:766825833740
另外，如果您使用Oracle，我会查看分析函数，特别是OVER PARTITION BY
答案 5 :(得分：1)
这个怎么样？
SELECT T1.f1 as "date 1", T2.f1 as "date 2", T3.f1 as "date 3"
FROM (SELECT *
FROM `date_test`
ORDER BY `f1` DESC
LIMIT 1) AS T1,
(SELECT *
FROM `date_test`
ORDER BY `f1` DESC
LIMIT 1, 1) AS T2,
(SELECT *
FROM `date_test`
ORDER BY `f1` DESC
LIMIT 2, 1) AS T3
;
哪个输出：
+------------+------------+------------+
| date 1 | date 2 | date 3 |
+------------+------------+------------+
| 2003-01-01 | 2002-01-01 | 2001-01-01 |
+------------+------------+------------+
唯一的缺点是你需要至少三行，否则它不会返回任何东西......
使用JOIN，您可以执行此操作：
SELECT T1.id,
T1.f1 as "date 1",
T2.f1 as "date 2",
T3.f1 as "date 3"
FROM `date_test` as T1
LEFT JOIN (SELECT * FROM `date_test` ORDER BY `f1` DESC) as T2 ON (T1.id=T2.id AND T1.f1 != T2.f1)
LEFT JOIN (SELECT * FROM `date_test` ORDER BY `f1` DESC) as T3 ON (T1.id=T3.id AND T2.f1 != T3.f1 AND T1.f1 != T3.f1)
GROUP BY T1.id
ORDER BY T1.id ASC, T1.f1 DESC
将返回类似的内容：
+----+------------+------------+------------+
| id | date 1 | date 2 | date 3 |
+----+------------+------------+------------+
| 1 | 2001-01-01 | 2003-01-01 | 2002-01-01 |
+----+------------+------------+------------+
缺点是date1，date 2和date 3不一定按特定顺序排列（根据上面的示例输出）。但这可以以编程方式实现。好的一面是，您可以在WHERE之前插入GROUP BY子句，例如，您可以按T1.id进行搜索。
答案 6 :(得分：0)
您可以通过旋转表格连续获得前3个日期，如果您希望它们在旋转后在一列中连接，也可以连接日期。
编辑： 这是一个用于透视表并连续提供最新3个日期的查询。但要进行透视，您需要知道表中可用的数据。我想，因为我们正在查询最新的3个日期，所以我们不知道在日期列中转动的确切值。首先，我将最新的3个日期查询到临时表中。然后在row_number 1,2和3上运行一个轴，以获得连续的最新3个日期。
Select Top 3 * into #Temp from Test order by f1 desc
现在，转向row_number列 -
SELECT id,[3] as Latest,[2] as LatestMinus1,[1] as LatestMinus2
FROM (
select ROW_NUMBER() OVER(ORDER BY f1) AS RowId,f1,id from #Temp) AS Src
PIVOT (Max(f1) FOR RowId IN ([1],[2],[3])) AS pvt
这导致 -
Id | Latest |LatestMinus1 |LatestMinus2
1 | 2003-01-01 00:00:00.000 | 2002-01-01 00:00:00.000 | 2001-01-01 00:00:00.000
当然还有
drop table #Temp