在每5行中选择最大值

我的总行数是可变的，并且不是固定的，所以有N行，我想将每5行分成一组，并在SQL中的下表中选择价格的最大值。在每5行中选择最大值

Date  Price 
20170101 100 
20170102 110 
20170103 90 
20170105 80 
20170109 76 

20170110 50 
20170111 55 
20170113 80 
20170115 100 
20170120 99 

20170121 88 
20170122 98 
2017

所以在第一组5最高价格是110，而第二组是100，而最后一组最高价格是120

来源

2017-08-14 MahdiIBM

使用公用表表达式对它们进行分组。

WITH CTE AS (SELECT RANK() OVER (ORDER BY Date) AS Rank, Price 
      FROM yourtable) 
SELECT (Rank - 1)/5 AS GroupedDate, MAX(Price) AS MAXPRICE 
FROM CTE 
GROUP BY ((Rank - 1)/5);

输出

GroupedDate MAXPRICE 
0   110 
1   100 
2   120

SQL小提琴：http://sqlfiddle.com/#!6/b5857/3/0

来源

2017-08-14 14:29:21 Matt

工作就像一个魅力，让我知道是有可能有日期选定的价格还？ – MahdiIBM

像在类似：'20170101 - 20170109 110' – Matt

不喜欢 - > 20170102 = 110 – MahdiIBM

你可以使用：

SELECT grp, MAX(Price) AS price 
FROM (SELECT *, ROW_NUMBER() OVER(ORDER BY DATE)/5 AS grp FROM tab) sub 
GROUP BY grp; 

-- OUTPUT 
grp price 
0 110 
1 100 
2 120

Rextester Demo

*假设日期是唯一的

编辑：

如像：20170101 - 20170109 110

SELECT 
    CONVERT(VARCHAR(8),MIN(DATE),112) + '-' + CONVERT(VARCHAR(8),MAX(date),112) 
    , MAX(Price) AS price 
FROM (SELECT *, (ROW_NUMBER() OVER(ORDER BY DATE))/5 AS grp FROM tab) sub 
GROUP BY grp;

输出：

20170101-20170105 110 
20170109-20170115 100 
20170120-2017

Rextester Demo2

来源

2017-08-14 14:27:57 lad2025

您可以使用ROW_NUMBER如下

;With cte as (
Select *, Bucket = Sum(RowN) over(Order by [date]) from (
    Select *, RowN = case when row_number() over(order by [date]) % 5 = 0 then 1 else 0 end from #data1 
) a 
) Select top (1) with ties [Date], [Price] 
from cte 
order by row_number() over (partition by Bucket order by Price desc)

来源

2017-08-14 14:36:56

在每5行中选择最大值

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