use*_*108 16 mysql mysql-workbench mysql-error-1111
这样可行:
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
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但是我只需要获取大于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
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那是我得到的时候:
错误代码1111.无效使用组功能
也就是说,当我在以下情况中添加此条件时,它会失败WHERE
:
AND ROUND(100*(SUM(ci.population)/c.population)) > 30
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Sur*_*rgy 30
所以你必须将这个条件移到HAVING子句中
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
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您在where子句中使用聚合函数,而SQL中无法做到这一点。
改用HAVING子句:
WHERE c.continent = 'Europe'
GROUP BY c.name
HAVING ROUND(100*(SUM(ci.population)/c.population)) > 30
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