SQL Group By Function

Question

我需要过滤包含多个位置的状态的数据。我的代码是：

SELECT
  cc.country_id,
  cc.country_name,
  l.city,
  l.street_address,
  l.postal_code,
  l.state_province,
  count(*)
  FROM locations l, countries cc
 WHERE l.country_id = cc.country_id
 ORDER BY cc.country_id
 GROUP BY (cc.country_id, cc.country_name)
HAVING count(*) > 1;

我收到以下消息：

ORA-00937: not a single-group group function
00937. 00000 -  "not a single-group group function"
*Cause:    
*Action:
Error at Line: 6 Column: 9

有什么问题？

Answer 1

所有不在聚合函数中的字段（count，sum，avg等）必须在group by子句中

那么，你应该有

group by cc.country_id, cc.country_name, 
         l.city, l.street_address, l.postal_code, l.state_province

或者你应该从select子句中删除这些字段，或者将它们放在select子句中的聚合函数中，或者更改你的查询。

顺便说一下，最好使用连接语法

替换

FROM locations l, countries cc
 where l.country_id = cc.country_id

通过

FROM location l
inner join countries cc on l.country_id = cc.country_id

最后，ORDER BY必须是查询中的最后一个语句

Answer 2

查询存在以下几个问题：

1. group by需要所有字段
2. order by位置错误
3. 它使用隐式而非显式join语法

以下内容应该有效：

 SELECT cc.country_id, cc.country_name, l.city, l.street_address,
        l.postal_code, l.state_province, count (*)
 FROM locations l join
      countries cc
      on l.country_id = cc.country_id
 group by cc.country_id, cc.country_name, l.city, l.street_address,
          l.postal_code, l.state_province
 having count(*) > 1
 order by cc.country_id;

但是，如果您想计算国家/地区中的行数，请删除其他字段：

 SELECT cc.country_id, cc.country_name, count (*)
 FROM locations l join
      countries cc
      on l.country_id = cc.country_id
 group by cc.country_id, cc.country_name
 having count(*) > 1
 order by cc.country_id;

Answer 3

您可以利用窗口函数

SELECT c.country_id, c.country_name, l.city, l.street_address, l.postal_code, l.state_province, l.l_count 
  FROM countries c JOIN 
(
  SELECT country_id, city, street_address, postal_code, state_province,
         COUNT(*) OVER (PARTITION BY country_id) l_count
    FROM locations
) l
    ON l.country_id = c.country_id
 WHERE l.l_count > 1
 ORDER BY c.country_id

这是 SQLFiddle 演示