Sql查询捕获最小日期，基于行排序的最大日期为空

Question

议程是跟踪资源的雇用和重新雇用方案

empID   emp_Status      CreateDate      end_date
1       active          **2019-02-01**  Null
1       active          2019-02-02      2019-01-04 (Skipped because last day is greater than joining date)
1       active          2019-02-03      Null
1       Terminated      2019-02-04      **2019-02-02**
1       Terminated      2019-02-05      2019-02-02
1       active          2019-02-06      Null

输出应该能够跟踪加入日期和最后一个工作日以及相应的计数

输出：

empID   join_date       last_date       Joining_count
1       2019-02-01      2019-02-02      1
1       2019-02-06      Null            2

我需要在 redshift 或 oracle sql 查询中实现这一点。请帮我解决这个问题。

Answer 1

如果我理解正确，对于每个“新”活动，您想要下一个“终止”。一种方法是创建反向计数的组，然后聚合：

select empid,
       min(case when emp_status = 'active' then createdate end) as active_date,
       min(case when emp_status = 'Terminated' then createdate end) as terminate_date,
       row_number() over (partition by empid order by min(createdate)) as joining_count
from (select t.*,
             sum(case when emp_status = 'Terminated' then 1 else 0 end) over (partition by empid order by createdate desc) as grp
      from t
     ) t
group by empid, grp
having sum(case when emp_status = 'active' then 1 else 0 end) > 0;

Here 是一个 dbfiddle。