错误代码1111。无效使用组功能

Posted 2023-02-16

【中文标题】错误代码1111。无效使用组功能

【英文标题】：Error Code 1111. Invalid use of group function

【发布时间】：2014-04-04 05:14:05

【问题描述】：

所以这行得通：

SELECT c.name AS country_name, c.population AS country_population, SUM(ci.population) AS city_population, ROUND(100*(SUM(ci.population)/c.population)) AS city_population_percent
FROM country AS c
JOIN city AS ci
ON c.code = ci.countrycode
WHERE c.continent = 'Europe'
GROUP BY c.name

但我只需要获取大于 30 的 city_population_percent 值，所以我试试这个：

SELECT c.name AS country_name, c.population AS country_population, SUM(ci.population) AS city_population, ROUND(100*(SUM(ci.population)/c.population)) AS city_population_percent
FROM country AS c
JOIN city AS ci
ON c.code = ci.countrycode
WHERE c.continent = 'Europe'
**AND ROUND(100*(SUM(ci.population)/c.population)) > 30**
GROUP BY c.name

那时我得到了：

错误代码1111。组函数使用无效

也就是说，当我在WHERE中添加这个条件时它失败了：

AND ROUND(100*(SUM(ci.population)/c.population)) > 30

【问题讨论】：

除了使用having 子句之外：您对group by 的一般用法在（标准）SQL 中是无效的。 mysql 对group by 的草率实现允许在任何其他DBMS 中失败的东西。有关详细信息，请阅读：mysqlperformanceblog.com/2006/09/06/… 和此：rpbouman.blogspot.de/2007/05/debunking-group-by-myths.html

【参考方案1】：

所以你必须把这个条件移到 HAVING 子句

SELECT c.name AS country_name, c.population AS country_population, SUM(ci.population) AS city_population, ROUND(100*(SUM(ci.population)/c.population)) AS city_population_percent
            FROM country AS c
            JOIN city AS ci
            ON c.code = ci.countrycode
WHERE c.continent = 'Europe'
GROUP BY c.name
HAVING ROUND(100*(SUM(ci.population)/c.population)) > 30

【讨论】：

不要忘记group by 的这种无效使用只会返回可能不是正确结果的随机数据。

【参考方案2】：

您在 where 子句中使用聚合函数，这在 SQL 中是做不到的。

改用HAVING 子句：

WHERE c.continent = 'Europe'
GROUP BY c.name
HAVING ROUND(100*(SUM(ci.population)/c.population)) > 30