Postgresql-选择相应的lead列

我可以通过使用找到每行的lead status date

select lead(status_date) OVER(ORDER BY dept_id, status_date, id) as "lead_date" from my_table

如何按照与潜在客户状态日期相同的顺序获取相应的潜在客户状态

id, dept_id, status, statud_date
---------------------------------
1   001      OPEN    1/1/2017
2   001      ACTIVE  2/2/2017
3   002      CLOSED  1/15/2017

---

将参数更改为

lead（）

：

通常，我希望按部门列出状态，因此我认为这是您想要的：

select lead(status_date) OVER (partition by dept_id ORDER BY status_date, id) as next_status_date,
       lead(status) OVER (partition by dept_id ORDER BY status_date, id) as next_status      
from my_table;

当多个窗口规范使用同一子句时，可以指定该子句一次：

select lead(status_date) over w as next_status_date,
       lead(status) over w as next_status      
from my_table
window w as (partition by dept_id order by status_date, id) ;

---

使用

status

列上的

lead

。

select lead(status_date) over w as next_status_date,
       lead(status) over w as next_status      
from my_table
window w as (partition by dept_id order by status_date, id) ;