你可以用 EXISTS :

SELECT lf.*
FROM LINKAGE_FILE lf
WHERE EXISTS (SELECT 1 FROM LINKAGE_FILE lf1 WHERE lf1.ACCOUNT_NUM = lf.ACCOUNT_NUM AND lf1.UNIQUE_CUSTOMER_ID <> lf.UNIQUE_CUSTOMER_ID);

但是，您也可以使用查询进行聚合：

SELECT ACCOUNT_NUM, COUNT(DISTINCT UNIQUE_CUSTOMER_ID) 
FROM LINKAGE_FILE   
GROUP BY ACCOUNT_NUM
HAVING COUNT(DISTINCT UNIQUE_CUSTOMER_ID) > 1;

这样，你只能得到 ACCOUNT_NUM 有两个或更多 CUSTOMER_ID s。

您只需使用 COUNT(DISTINCT..) 每 account_num 并过滤掉计数大于1的情况，在 HAVING 条款：

SELECT
  ACCOUNT_NUM, 
  COUNT(DISTINCT CUSTOMER_ID) AS unique_customer_count 
FROM LINKAGE_FILE 
GROUP BY ACCOUNT_NUM
HAVING unique_customer_count > 1

将客户支票放入一个join查询中，如下所示

DROP TABLE if exists t;
create table t(accountid int,cid int);
insert into t values
(1,1),(1,2).(1,1),(2,3),(3,4),(3,4);

    select distinct t.accountid,t.cid 
from t
join
(
select accountid,count(distinct cid) cids
from t
group by accountid having cids > 1 
) s on s.accountid = t.accountid;

+-----------+------+
| accountid | cid  |
+-----------+------+
|         1 |    1 |
|         1 |    2 |
+-----------+------+
2 rows in set (0.00 sec)