sql group by columns

Question

我有一张桌子：

ID    Status    Date
1     Stop      01.01.2001
1     Run       02.01.2002
2     Stop      03.01.2003
2     Run       04.01.2004

我想按如下方式对行进行分组：

ID    Stop         Run    
1     01.01.2001   02.01.2002
2     03.01.2003   04.01.2004

制作这个小组的最佳方法是什么？

Answer 1

select id,
       max(case when status = 'Stop' then date end) as stop,
       max(case when status = 'Run' then date end) as run
from your_table
group by id

Answer 2

使用以下查询来使用

获取输出

CASE和GROUP BY

select ID,
 MIN(case when action = 'Stop' then dateval end) STOP,
 MIN(case when action = 'Run' then dateval end) RUN
from my_table
group by id;

要使用GROUP BY，您需要使用一些聚合函数。因此，您可以使用MIN或MAX

---

查询测试数据：

with my_table as
(
    select 1 id,      'Stop' action,       '01.01.2001' dateval from dual
    union 
    select 1 id,      'Run' action,       '02.01.2002' dateval from dual
    union 
    select 2 id,      'Stop' action,       '03.01.2003' dateval from dual
    union 
    select 2 id,      'Run' action,       '04.01.2004' dateval from dual
)
select ID,
 MIN(case when action = 'Stop' then dateval end) STOP,
 MIN(case when action = 'Run' then dateval end) RUN
from my_table
group by id

Answer 3

使用

ORACLE的PIVOT功能

，可以使用以下查询完成相同的操作：

SELECT *
FROM   (SELECT ID,dateval,action
        FROM   my_table)
PIVOT  (max(dateval) FOR (action) IN ('Stop' AS STOP, 'Run' AS RUN));

---

查询测试数据：

with my_table as
(
select 1 id,      'Stop' action,       '01.01.2001' dateval from dual
union 
select 1 id,      'Run' action,       '02.01.2002' dateval from dual
union 
select 2 id,      'Stop' action,       '03.01.2003' dateval from dual
union 
select 2 id,      'Run' action,       '04.01.2004' dateval from dual
)

SELECT *
FROM   (SELECT ID,dateval,action
        FROM   my_table)
PIVOT  (max(dateval) FOR (action) IN ('Stop' AS STOP, 'Run' AS RUN));