Question

我在oracle中有2个表名为employee_1和mobile_1 empolyee_1数据

  Mobile_No  Sim_No      Start_Date         End_Date 
1111111113  1111112222  11/10/2017 21:02:44 13/10/2017 21:02:44
1111111111  1212121212  3/10/2017 21:02:44  10/10/2017 21:02:44
1111111111  1111111111  11/10/2017 21:02:44 13/10/2017 21:02:44
1111111111  1111111112  11/10/2017 21:02:44 13/10/2017 21:02:44
1111111111  1111111111  12/10/2017 21:02:44

Mobile_1数据

  Mobile_No  Status
1111111111  connected
1111111112  connected

我想检索第一个表的所有4列。条件是empolyee_1.mobile_No = Mobile_1.mobile_no，具有最长开始日期。

Answer 1

以下是解决方案，请查看...

select mobile_no,sim_no,start_date,end_date from(
select mobile_no,sim_no,start_date,end_date,rank() over(partition by mobile_no,sim_no order by start_date desc) rn from employee_1)s
where rn=1
and exists (select 1 from mobile_1 m where m.Mobile_No = s.mobile_no);

Answer 2

可以使用

INNER JOIN：

select e1.* 
  from employee_1 e1 
  inner join ( select mobile_no, max(e.start_date) start_date from employee_1 e group by mobile_no ) e2 
          on ( e1.mobile_no = e2.mobile_no and e1.start_date = e2.start_date )
  inner join mobile_1 m
          on ( m.mobile_no = e1.mobile_no );

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Oracle查询聚合函数

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