我现在要查的是top client(买东西总和加起来最多的)
所以我输了 SELECT C.CLIENTNO, C.CNAME, MAX(SUM(P.AMOUNT))
FROM CLIENT C,PURCHASE P
WHERE C.CLIENTNO=P.CLIENTNO
GROUP BY C.CLIENTNO,C.CNAME;
提示说不是单组分组函数,所以 问题应该是出在 max(sum(p.amount))上吧,应该如何分开来写啊
------解决方案--------------------
with t as
(SELECT job, deptno, sum(sal) sal FROM emp group by job, deptno)
select * from t where sal = (select max(sal) from t)
------解决方案--------------------
SELECT CLIENTNO,CNAME,MAX_VALUE
FROM
(SELECT C.CLIENTNO, C.CNAME, SUM(P.AMOUNT) MAX_VALUE
FROM CLIENT C,PURCHASE P
WHERE C.CLIENTNO=P.CLIENTNO
GROUP BY C.CLIENTNO,C.CNAME) TT1
WHERE TT1.MAX_VALUE
IN
(SELECT MAX(MAX_VALUE)
FROM
(SELECT C.CLIENTNO, C.CNAME, SUM(P.AMOUNT) MAX_VALUE
FROM CLIENT C,PURCHASE P
WHERE C.CLIENTNO=P.CLIENTNO
GROUP BY C.CLIENTNO,C.CNAME) TT);