使用 GROUP BY 和 HAVING COUNT(*) >1 到 select 重复和非重复字段

Question

下面是CUST_REFtable

中的数据

CUST ID REF ID 1 7 2 2 3 5 4 5 5 5

下面的查询将 return 3 5 和 4 5

SELECT CUST_ID, REF_ID
FROM CUST_REF
WHERE REF_ID IN 
(select ref_id from CUST_REF
 group by ref_id
 having (count(*) > 1))
AND CUST_ID != REF_ID;

如果想return1 72 25 5怎么样？我在下面查询它只能 return 1 7 和 2 2

SELECT CUST_ID, REF_ID
FROM CUST_REF
WHERE CUST_ID = REF_ID
AND REF_ID NOT IN 
(select ref_id from CUST_REF
group by ref_id
having (count(*) > 1))
UNION
SELECT CUST_ID, REF_ID
FROM CUST_REF
WHERE CUST_ID != REF_ID
AND REF_ID NOT IN 
(select ref_id from CUST_REF
group by ref_id
having (count(*) > 1));

Answer 1

您似乎想要 select REF_ID 不重复的行，以及 REF_ID = CUST_ID 的行。如果是这样，这个怎么样？

SQL> with cust_ref(cust_id, ref_id) as
  2  (select 1, 7 from dual union
  3   select 2, 2 from dual union
  4   select 3, 5 from dual union
  5   select 4, 5 from dual union
  6   select 5, 5 from dual
  7  )
  8  select cust_id, ref_id
  9  from cust_ref
 10  where ref_id not in (select ref_id
 11                       from cust_ref
 12                       group by ref_id
 13                       having count(*) > 1
 14                      )
 15     or ref_id = cust_id;

   CUST_ID     REF_ID
---------- ----------
         1          7
         2          2
         5          5

SQL>

使用 GROUP BY 和 HAVING COUNT(*) >1 到 select 重复和非重复字段

Using GROUP BY and HAVING COUNT(*) >1 to select duplicate and non-duplicate field

oracle

group-by

count

having-clause