如何从表中获得几个最高值？

Question

我有这样

id f1 
-------------- 
1 2000-01-01 
1 2001-01-01 
1 2002-01-01 
1 2003-01-01 

一个表，我想说，最近3日在一排

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 

试图像

  SELECT T1.*,T2.* 
      FROM Test AS T1 
LEFT OUTER JOIN Test AS T2 ON T1.id = T2.id AND (T2.f1 > T1.f1) 

Answer 1

在SQL Server中，你可以做select top 3 * from Test order by f1 desc。其他DBMS的也有类似的posibilities如MySQL的limit，Oracle的rownum等

Answer 2

东西虽然我不是当然如何让他们成为一排，你可以从下面开始：

SELECT * FROM Test ORDER BY f1 DESC LIMIT 3 

这应该给你一个结果，如：

id f1 
1 2003-01-01 
1 2002-01-01 
1 2001-01-01 

将它们放入单列，虽然可能有点难度......

Answer 3

您可以用ORDER BY，TOP和PIVOT组合，至少在SQL Server上做到这一点。似乎许多其他答案已经忽略了结果是“全部在一行上”的需要。

Answer 4

如果您希望在透视后旋转一列，则可以通过旋转该表并连接日期来获得连续排名前3的日期。

编辑： 这是一个查询来转换表并提供连续的最新3个日期。但是为了支撑你，你需要知道表格中可用的数据。我想，因为我们正在查询最新的3个日期，所以我们不会知道在日期列中转换的确切值。所以首先，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上运行一个数据透视表以获取连续的最新3个日期。

Select Top 3 * into #Temp from Test order by f1 desc 

现在，在ROW_NUMBER列支点 -

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

drop table #Temp 

Answer 5

什么你正在尝试做的被称为数据透视表，这里是一篇关于如何的文章做到这一点：

http://asktom.oracle.com/pls/asktom/f?p=100:11:0::NO::P11_QUESTION_ID:766825833740

另外，如果您使用Oracle我想看看解析函数，OVER PARTITION BY特别

Answer 6

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 
; 

，输出：

+------------+------------+------------+ 
| date 1  | date 2  | date 3  | 
+------------+------------+------------+ 
| 2003-01-01 | 2002-01-01 | 2001-01-01 | 
+------------+------------+------------+ 

唯一的缺点是，你至少需要三人行，否则它不会返回任何东西......

使用JOIN，你可以这样做：

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 

这将返回类似于：

+----+------------+------------+------------+ 
| 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不一定会以特定的顺序（按照上面的示例输出）。但这可以通过编程实现。有利的一面是，你可以在GROUP BY前插入一个WHERE条款，您可以通过搜索T1.id，例如。

在T-SQL

Answer 7

（这将让你的前三名日期，即使他们都是相同的值）

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] 

这将让你顶三个不同的日期

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] 

尝试这在SQL Server 2005中

--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