Question

我有一张这样的桌子：

state         date
new           01/01/2018
new           02/01/2018
active        05/01/2018
active        09/01/2018
active        10/01/2018
new           12/01/2018
active        13/01/2018
active        14/01/2018
close         15/01/2018

每个状态的最小日期，我需要在状态栏中进行任何更改。

决赛桌：

state         date
new          01/01/2018
active       05/01/2018
new          12/01/2018
close        15/01/2018

如何在SQL中做到这一点？

Answer 1

由于您尚未标记任何DBMS，所以我会选择row_number()，因为它似乎是gaps-and-islands的问题：

select state, min(date)
from (select t.*,
             row_number() over (order by date) as seq1,
             row_number() over (partition by state order by date) as seq2
      from table t
      ) t
group by state, (seq1-seq2);

Answer 2

这是使用WITH和LAG的解决方案

WITH state_change (state, st_date, next_state) AS
  (SELECT state , st_date, 
          LAG(state, 1, '') OVER(ORDER BY st_date) as s2
   FROM table
 )
SELECT state, st_date
FROM state_change
WHERE state != next_state
ORDER BY st_date ASC

输出（包括激活的第二个更改）

state   st_date
new     2018-01-01
active  2018-01-05
new     2018-01-12
active  2018-01-14
close   2018-01-15

我如何选择列中的任何更改？

2 个答案: