如何序列号子组

时间:2013-08-29 12:01:02

标签: sql-server-2008

我有一个像下面的表结构

id  wstage      wstatus wdate
101 Unaquired   create  2013-08-29 17:07:20.040
101 Unaquired   rework  2013-08-29 18:07:20.040
101 inprocess   accqui  2013-08-29 19:07:20.040
101 inprocess   alloca  2013-08-29 20:07:20.040
101 Unaquired   create  2013-08-29 21:07:20.040
101 Unaquired   rework  2013-08-29 22:07:20.040

我必须将此编号为

id  wstage      wstatus wdate                   rownumber
101 Unaquired   rework  2013-08-29 22:07:20.040 1
101 Unaquired   create  2013-08-29 21:07:20.040 1
101 inprocess   alloca  2013-08-29 20:07:20.040 2
101 inprocess   accqui  2013-08-29 19:07:20.040 2
101 Unaquired   rework  2013-08-29 18:07:20.040 3
101 Unaquired   create  2013-08-29 17:07:20.040 3

我正在尝试使用功能

select *,ROW_NUMBER() over (partition by id,wstage order by wdate desc) rownumber

但这并没有给出理想的输出。我不想使用pl / sql是否有排名函数或简单查询来实现这一点。我的桌子有5000万条记录。

1 个答案:

答案 0 :(得分:4)

假设每个wdate id值是唯一的,这可能会起作用:

WITH partitioned AS (
  SELECT
    *,
    grp = ROW_NUMBER() OVER (PARTITION BY id         ORDER BY wdate DESC)
        - ROW_NUMBER() OVER (PARTITION BY id, wstage ORDER BY wdate DESC)
  FROM atable
),
maxdates AS (
  SELECT
    id, wstage, wstatus, wdate,
    maxwdate = MAX(wdate) OVER (PARTITION BY id, wstage, grp)
  FROM partitioned
)
SELECT
  id, wstage, wstatus, wdate,
  rownumber = DENSE_RANK() OVER (PARTITION BY id ORDER BY maxwdate DESC)
FROM maxdates
;

第一个CTE确定不同的id, wstage个岛,第二个CTE找到每个岛的最大wdate,主查询根据找到的最大值对行进行排名。

此查询的SQL小提示演示is available