如何从 creation_time 获取 start_date 和 end_date

Question

需要帮助，我需要根据 is_active state 从 creation_time 获取 start_date 和 end_date。 我尝试了几个查询，但没有得到正确的结果。

表格示例

ID	用户身份	姓名	领导者姓名	活跃	创建时间
6	29	东风	作为	0	2021-10-10
620	29	东风	RB	0	2022-02-09
1088	29	东风	作为	1	2022-06-30

结果应该是这样的：

ID	用户身份	姓名	领导者姓名	活跃	开始日期	结束日期	创建时间
6	29	东风	作为	0	2021-10-10	2022-02-09	2021-10-10
620	29	东风	RB	0	2022-02-09	2022-06-30	2022-02-09
1088	29	东风	作为	1	2022-06-30	当前的日期（）	2022-06-30

请各位朋友帮忙，先谢谢了

Answer 1

根据问题部分和评论部分的信息，我相信 is_active=1 的行具有组的最新创建时间（基于 user_id）。这是在工作台中编写和测试的查询。

select id,user_id,name,leader_name,is_active,
t1.creation_time as start_date, case is_active when 0 then t2.creation_time else current_date() end as end_date,t1.creation_time
from (select id,user_id,name,leader_name,is_active,creation_time,@row_id:=@row_id+1 as row_id
    from test,(select @row_id:=0)t
    where user_id=29
    order by creation_time
    )t1
left join
    (select creation_time,@row_num:=@row_num+1 as row_num
    from test,(select @row_num:=0)t
    where user_id=29
    order by creation_time
    )t2
on t1.row_id+1=t2.row_num
;

-- result set:
# id, user_id, name, leader_name, is_active, start_date, end_date, creation_time
6, 29, DF, AS, 0, 2021-10-10, 2022-02-09, 2021-10-10
620, 29, DF, RB, 0, 2022-02-09, 2022-06-30, 2022-02-09
1088, 29, DF, AS, 1, 2022-06-30, 2022-08-31, 2022-06-30

这还不是结束。 以防万一您想根据每个 user_id 组显示 output，下面是代码：

-- first of all insert the following 4 lines on top of the original table data, which has the same user_id 50 
61  50  DF  AS  0   2021-10-10
630 50  DF  RB  0   2022-02-09
1188    50  DF  TS  0   2022-06-30
2288    50  DF  AS  1   2022-07-30

select  id,t1.user_id,name,leader_name,is_active,
t1.creation_time as start_date, case is_active when 0 then t2.creation_time else current_date() end as end_date,t1.creation_time
from 
 (select id,user_id,name,leader_name,is_active,creation_time,@row_id:=@row_id+1 as row_id
    from test,(select @row_id:=0)t
    order by user_id,creation_time
    )t1
left join
    (select user_id,creation_time,@row_num:=@row_num+1 as row_num
    from test,(select @row_num:=0)t
    order by user_id,creation_time
    )t2
on t1.user_id=t2.user_id and t1.row_id+1=t2.row_num
;

-- result set:
# id, user_id, name, leader_name, is_active, start_date, end_date, creation_time
6, 29, DF, AS, 0, 2021-10-10, 2022-02-09, 2021-10-10
620, 29, DF, RB, 0, 2022-02-09, 2022-06-30, 2022-02-09
1088, 29, DF, AS, 1, 2022-06-30, 2022-08-31, 2022-06-30
61, 50, DF, AS, 0, 2021-10-10, 2022-02-09, 2021-10-10
630, 50, DF, RB, 0, 2022-02-09, 2022-06-30, 2022-02-09
1188, 50, DF, TS, 0, 2022-06-30, 2022-07-30, 2022-06-30
2288, 50, DF, AS, 1, 2022-07-30, 2022-08-31, 2022-07-30