我想在Oracle表中选择非重复的记录（只选择重复记录）

Question

以下是表（客户），并记录如下：

CUSTID CUSTNAME 
100  ashwin 
100  ashwin 
200  adithyan 
100  ashwin 
300  aravind 
300  aravind 
300  aravind 

我想要的结果是这样（非重复行）：

CUSTID CUSTNAME 
100  ashwin 
100  ashwin 
300  aravind 
300  aravind 

Answer 1

试试这个：

select custid, custname 
from customer a where 
exists (
select CUSTID, CUSTNAME, count(*) 
from customer b 
where a.custid = b.custid and a.custname = b.custname 
group by CUSTID, CUSTNAME 
having count(*) > 1) 

Answer 2

我想要的结果是这样的（非重复行）：

CUSTID CUSTNAME 
100  ashwin 
100  ashwin 
300  aravind 
300  aravind 

如果我理解正确的话，你要重复行的其余部分。您可以使用分析ROW_NUMBER。

测试用例：

SQL> CREATE TABLE t 
    2  (CUSTID int, CUSTNAME varchar2(8)); 

Table created. 

SQL> 
SQL> INSERT ALL 
    2  INTO t (CUSTID, CUSTNAME) 
    3   VALUES (100, 'ashwin') 
    4  INTO t (CUSTID, CUSTNAME) 
    5   VALUES (100, 'ashwin') 
    6  INTO t (CUSTID, CUSTNAME) 
    7   VALUES (200, 'adithyan') 
    8  INTO t (CUSTID, CUSTNAME) 
    9   VALUES (100, 'ashwin') 
10  INTO t (CUSTID, CUSTNAME) 
11   VALUES (300, 'aravind') 
12  INTO t (CUSTID, CUSTNAME) 
13   VALUES (300, 'aravind') 
14  INTO t (CUSTID, CUSTNAME) 
15   VALUES (300, 'aravind') 
16 SELECT * FROM dual; 

7 rows created. 

SQL> 
SQL> COMMIT; 

Commit complete. 

SQL> 

查询使用ROW_NUMBER

SQL> select custid, custname 
    2 from (
    3 SELECT custid, custname, 
    4   row_number() over (partition by custid order by custid) as rn 
    5 from t 
    6 ) 
    7 WHERE rn > 1 
    8 order by custid; 

    CUSTID CUSTNAME 
---------- -------- 
     100 ashwin 
     100 ashwin 
     300 aravind 
     300 aravind 

SQL> 

使你所需的输出。