Question

当您更改一组empID，DepID时，结果列应增加1。如果连续2行的某些DT之间的天数大于5天，则结果列也应重置。

libs

Answer 1

我想你想要row_number()。但是，大于5天的情况＆＃34;很棘手您还需要识别行组。为此，使用lag()查找组开始的时间。然后是累计和来指定组，最后是row_number()：

select EmpID, DepID, someDt,
       row_number() over (partition by EmpID, DepId, grp, order by someDt) as Result
from (select t.*,
             sum(case when someDt < prevDt + 5 then 0 else 1 end) over
                 (partition by EmpID, DepId order by someDt) as grp
      from (select t.*,
                   lag(someDt) over (partition by EmpID, DepId order by someDt) as prevDt
            from t
           ) t
     ) t;

Answer 2

我可能觉得太复杂了。嗯，这是我的方法：

检测所需的群组中断记录;用1标记它们，而其他记录得到0
在该分组中断号上有一个总计，因此您会得到编号组
为每组记录编号。

以下是查询：

select 
  empid,
  depid,
  somedt, 
  row_number() over (partition by grp order by empid, depid, somedt) as result
from
(
  select 
    empid, depid, somedt, sum(chg) over (order by empid, depid, somedt) as grp
  from
  (
    select 
      empid, depid, somedt,
      case when empid = lag(empid) over (order by empid, depid, somedt)
            and depid = lag(depid) over (order by empid, depid, somedt)
            and somedt <= lag(somedt) over (order by empid, depid, somedt) + 5
           then 0
           else 1
      end as chg
    from mytable
  )
);

我想要一个查询来生成下面的结果列。假设所需的列存在于同一个表中

2 个答案: