使用 SQL 中的 RANK() 作为记录组的 ID 号

Using RANK() in SQL as ID Number for Groups of Records

这是我的 table:

employeeid  workdate                workstatus
----------- ----------------------- ----------
1           2020-09-01 00:00:00.000 ON 
1           2020-09-02 00:00:00.000 ON 
1           2020-09-03 00:00:00.000 ON 
1           2020-09-04 00:00:00.000 OFF
1           2020-09-05 00:00:00.000 OFF
2           2020-09-01 00:00:00.000 ON 
2           2020-09-02 00:00:00.000 ON 
2           2020-09-03 00:00:00.000 OFF
2           2020-09-04 00:00:00.000 OFF
2           2020-09-05 00:00:00.000 ON 

我正在执行这个查询:

select employeeid, workdate, workstatus, rank() over(partition by employeeid, workstatus order by workdate) as cycle
from #workstatus
order by 1, 2

结果如下:

employeeid  workdate                workstatus cycle
----------- ----------------------- ---------- --------------------
1           2020-09-01 00:00:00.000 ON         1
1           2020-09-02 00:00:00.000 ON         2
1           2020-09-03 00:00:00.000 ON         3
1           2020-09-04 00:00:00.000 OFF        1
1           2020-09-05 00:00:00.000 OFF        2
2           2020-09-01 00:00:00.000 ON         1
2           2020-09-02 00:00:00.000 ON         2
2           2020-09-03 00:00:00.000 OFF        1
2           2020-09-04 00:00:00.000 OFF        2
2           2020-09-05 00:00:00.000 ON         3

我的目标是让 on/off 工作的“周期”由每个员工的唯一编号来标识。因此,员工 1 的三个工作日将是周期 1,然后两个休息日将是周期 2。

员工 2 的前两天 ON 日将是第 1 个周期,然后这两个 OFF 日将是第 2 个周期,最后一个 ON 日将是第 3 个周期。

我不确定是否可以为此使用 RANK(),或者是否有更好的解决方案。谢谢!

这是一种 gaps-and-islands 问题。对于此版本,使用 lag() 和累积总和:

select t.*,
       sum(case when prev_ws= workstatus then 0 else 1 end) over 
           (partition by employeeid order by workdate) as ranking
from (select t.*, 
             lag(workstatus) over (partition by employeeid order by workdate) as prev_ws
      from t
     ) t;

使用dense_rank代替排名

您可以使用 window 函数来解决这个 gaps-and-islands 问题。一种方法是利用行号之间的差异来构建“相邻”记录组:

select employeeid, workdate, workstatus,
    row_number() over(partition by employeeid, workstatus, rn1 - rn2 order by workdate) cycle
from (
    select t.*,
        row_number() over(partition by employeeid order by workdate) rn1,
        row_number() over(partition by employeeid, workstatus order by workdate) rn2
    from mytable t
) t