SQL GROUP BY 和 HAVING

Question

所以我不会再遇到这个错误。

 select distinct substr(CUSTZIP, 1,5), AVG(CUSTBAL), custcity, custstate
 from customer
 group by CUSTCITY, custstate
 having CUSTSTATE = 'wa' AND avg(CUSTBAL) >100;

错误显示“不是 GROUP BY 表达式”，并建议将“substr(CUSTZIP, 1,5), AVG(CUSTBAL)”添加到 group by 子句中，但这也不起作用。我想要做的是仅列出西澳州各城市的邮政编码和余额平均值，并且余额超过 100 美元。有人可以帮我指出我的错误吗？我确信它很简单，但我似乎无法以我的初学者水平得到它。

Answer 1

问题不在于 having 子句。它是 substr(CUSTZIP, 1, 5)。这是解决该问题的一种方法:

select substr(CUSTZIP, 1, 5), AVG(CUSTBAL), custcity, custstate
from customer
group by CUSTCITY, custstate, substr(CUSTZIP, 1, 5)
having CUSTSTATE = 'wa' AND avg(CUSTBAL) > 100;

顺便说一下，group by 几乎不需要 select unique。

或者，您可以使用聚合函数:

select max(substr(CUSTZIP, 1, 5)), AVG(CUSTBAL), custcity, custstate
from customer
group by CUSTCITY, custstate
having CUSTSTATE = 'wa' AND avg(CUSTBAL) > 100;