我正在制作考勤系统。我将员工的交易存储在下表中:
我想为每个员工获取最早和最新的交易,包括 date 和 type 。
我可以使用分组和聚合获取日期。但是,我无法弄清楚如何使用它们获取类型。
请你帮帮我。
谢谢。
答案 0 :(得分:6)
这就是FIRST和LAST聚合函数的设计目标。
以下是文档的链接:
第一:http://download.oracle.com/docs/cd/E11882_01/server.112/e17118/functions065.htm#SQLRF00641 最后:http://download.oracle.com/docs/cd/E11882_01/server.112/e17118/functions083.htm#sthref1206
这是一个例子:
SQL> create table my_transactions (id,employee_id,action_date,type)
2 as
3 select 1, 1, sysdate, 'A' from dual union all
4 select 2, 1, sysdate-1, 'B' from dual union all
5 select 3, 1, sysdate-2, 'C' from dual union all
6 select 4, 1, sysdate-3, 'D' from dual union all
7 select 5, 2, sysdate-11, 'E' from dual union all
8 select 6, 2, sysdate-12, 'F' from dual union all
9 select 7, 2, sysdate-13, 'G' from dual
10 /
Table created.
SQL> select *
2 from my_transactions
3 order by id
4 /
ID EMPLOYEE_ID ACTION_DATE T
---------- ----------- ------------------- -
1 1 04-07-2011 10:15:07 A
2 1 03-07-2011 10:15:07 B
3 1 02-07-2011 10:15:07 C
4 1 01-07-2011 10:15:07 D
5 2 23-06-2011 10:15:07 E
6 2 22-06-2011 10:15:07 F
7 2 21-06-2011 10:15:07 G
7 rows selected.
SQL> select employee_id
2 , min(action_date) min_date
3 , max(type) keep (dense_rank first order by action_date) min_date_type
4 , max(action_date) max_date
5 , max(type) keep (dense_rank last order by action_date) max_date_type
6 from my_transactions
7 group by employee_id
8 /
EMPLOYEE_ID MIN_DATE M MAX_DATE M
----------- ------------------- - ------------------- -
1 01-07-2011 10:15:07 D 04-07-2011 10:15:07 A
2 21-06-2011 10:15:07 G 23-06-2011 10:15:07 E
2 rows selected.
的问候,
罗布。
答案 1 :(得分:3)
您可以尝试使用analytical(or windowing functions)
select *
from
(select id, employee_id, action_date,type,
max(action_date) over (partition by employee_id) max_action_date,
min(action_date) over (partition by employee_id) min_action_date
from transaction)
where action_date in (max_action_date, min_action_date)