Question

我在尝试使用（union / intersect / minus）计算主键ACNUM在这些表中的次数时遇到了一些困难。

我尝试了多种使用SET运算符的方法，但似乎无法弄明白。

即使是这样的事情似乎也不起作用：

SELECT results.ACNUM, COUNT(results.books), COUNT(results.interests)
FROM (
 SELECT ACNUM FROM ACADEMIC A
 UNION
 SELECT COUNT(PANUM) as books FROM AUTHOR B
 UNION
 SELECT COUNT(FIELDNUM) as interests FROM INTEREST C
) results;

基本上，我需要将此查询（有效）转换为上面的查询，我使用set运算符。

SELECT DISTINCT ACNUM
FROM ACADEMIC A
WHERE
  (SELECT COUNT(PANUM)
  FROM AUTHOR
  WHERE ACNUM = A.ACNUM) < 5
AND
  (SELECT COUNT(FIELDNUM)
  FROM INTEREST
  WHERE ACNUM = A.ACNUM) > 3;

Answer 1

我不知道为什么你坚持使用集合运算符，但INTERSECT将返回预期结果：

SELECT ACNUM
FROM ACADEMIC A

INTERSECT

SELECT ACNUM
FROM AUTHOR
GROUP BY ACNUM 
HAVING COUNT(PANUM) < 5

INTERSECT

SELECT ACNUM
FROM INTEREST
GROUP BY ACNUM 
HAVING COUNT(FIELDNUM) > 3;

Answer 2

您以错误的方式使用了 UNION 查询，因为Ist查询中的内部查询只生成了一列，而您从中获取了三列。

你可以在这里使用join，如下所示：

 select a.ACNUM, count(ar.PANUM), count(i.FIELDNUM)

 from ACADEMIC a join AUTHOR ar on a.ACNUM=ar.ACNUM
     join INTEREST i on a.ACNUM=i.ACNUM

 group by a.ACNUM 
 having count(ar.PANUM)<5 and count(i.FIELDNUM)>3;

Answer 3

 SELECT X.ACNUM, Y.books, Z.interests
 FROM (
 (SELECT ACNUM FROM ACADEMIC A)X
 LEFT JOIN
 (SELECT COUNT(PANUM) as books FROM AUTHOR B)Y
 ON X.ACNUM=Y.ACNUM
 LEFT JOIN
 (SELECT COUNT(FIELDNUM) as interests FROM INTEREST C)Z
 ON X.ACNUM=Z.ACNUM ) WHERE Y.books<5 and Z.interests>3;

Oracle DB Set运算符用于计算重复字段

3 个答案: