Question

I m 采用Oracle 10g。

如果有以下重复浏览()。 Same Employee with two Employee number:

Employee_No      Employee_Name      ID_NO
----------------------------------------------
0002345          John Debb          100345642
0030988          John Debb          100345642
----------------------------------------------

i 想取得以下结果:

Employee_No_1      Employee_No_2     Employee Name    ID_NO
----------------------------------------------------------------
0002345            0030988           John Debb        100345642
----------------------------------------------------------------

www.un.org/Depts/DGACM/index_spanish.htm 能否在结构上做到? 或者需要PL/SQL? 问题是什么?

Answer 1

所要求的格式并不大,但将处理可能只重复2件的情况。

SELECT e.Employee_No, e.Employee_Name, e.ID_NO
    FROM (SELECT Employee_Name, ID_NO
              FROM Employee
              GROUP BY Employee_Name, ID_NO
              HAVING COUNT(*) > 1) q
        INNER JOIN Employee e
            ON q.Employee_Name = e.Employee_Name
                AND q.ID_NO = e.ID_NO
    ORDER BY e.Employee_Name, e.ID_NO, e.Employee_No

Answer 2

SELECT MIN(Employee_no), MAX(employee_no), Employee_name, id_no
FROM Employee
GROUP BY Employee_name, id_no
HAVING MIN(employee_no) <> MAX(employee_no)

我没有做甲骨文,但我认为这只是应当工作的普通辛子。

Answer 3

学历如下:

select e1.employee_no, e2.employee_no, e1.employee_name, e1.id_no
from employee e1
join employee e2
    on e1.id_no = e2.id_no
where e1.employee_no < e2.employee_no

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