Question

我的桌子

id           updateddate        updatedbyuser   department
-------------------------------------------------------
40715449759  12/7/2018 12:24            1260    IB
40715449759  12/8/2018 10:40            1203    ESCALATION
40715449759  12/8/2018 10:40            task
40715449759  12/9/2018 9:58           client
40715449759  12/9/2018 12:09       300050282    ESCALATION
42058447691  12/23/2018 16:44           1712    TASK
42058447691  12/26/2018 19:33           4700    ESCALATION
42058447691  12/27/2018 17:26           7357    ESCALATION

输出

id            first_action_date
--------------------------------
40715449759   12/8/2018 10:40
42058447691   12/26/2018 19:33

我正在尝试从mytable提取id和first_action_time，其中updatedate的第一个实例是first_action_date，部门是ESCALATION以获取唯一ID

Answer 1

只需使用min作为

的聚合

select id, 
       min(first_action_date) 
       as first_action_date
  from mytable
 where department = 'ESCALATION'
 group by id

Answer 2

Postgres具有一个称为distinct on的功能，可以在这种情况下使用：

select distinct on (id) id, first_action_date
from t
where department = 'ESCALATION'
order by id, first_action_date asc;

尽管您的特定查询可以直接由group by处理，但是如果您想要其他列（例如用户），这将很方便。要获取整行：

select distinct on (id) t.*
from t
where department = 'ESCALATION'
order by id, first_action_date asc;

Answer 3

select id, updateddate as first_action_date from table where department= 'ESCALATION' order by id

如何使用SQL查询获取输出列

3 个答案: