Question

如果问题很愚蠢，我很抱歉，因为我是SQL Server的新手。我想为指定月份的每一天选择前5个记录。例如

top 5 records for day 1 in month september
top 5 records for day 2 in month september
top 5 records for day 3 in month september
.
.
top 5 records for day 31 in month september

并将所有记录显示为一个结果。

Answer 1

假设你正在检查2012年6月的超速记录，你想要前5个速度（按速度desc）。

SELECT *
  FROM (
    SELECT *, RowNum = Row_number() over (partition by Cast(EventTime as Date)
                                          order by Speed desc)
      FROM Events
     WHERE EventTime >= '20120601'
       AND EventTime <  '20120701'
  ) X
 WHERE RowNum <= 5

Answer 2

试试这个，

WITH TopFiveRecords
AS
(
    SELECT  *,
            ROW_NUMBER() OVER (PARTITION BY dayColumn ORDER BY colName DESC) RN
    FROM tableName
)
SELECT  *
FROM TopFiveRecords
WHERE RN <= 5 
      -- AND date condition here ....

dayColumn包含月份日期的列
colName要排序的列

选择最高值对于指定月份的每一天

2 个答案: