问题描述
这是我的桌子：
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
我正在执行此查询：
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
我的目标是通过每个员工的唯一编号来标识开/关工作的“周期”。因此，员工1的三个工作日为周期1，那么两个工作日为周期2。
员工2的前两个工作日为第1个周期，然后这两个工作日为第2个周期，最后一个工作日为第3个周期。
我不确定是否可以为此使用RANK（）或是否有更好的解决方案。谢谢！
解决方法
这是一种空白和岛屿问题。对于此版本，请使用lag()和累计金额：
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density_rank代替等级
,您可以使用窗口功能来解决此“空缺”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,rn1 - rn2 order by workdate) cycle
from (
select t.*,row_number() over(partition by employeeid order by workdate) rn1,workstatus order by workdate) rn2
from mytable t
) t