聚合函数是否需要 GROUP BY？

Question

问题：

Given the CITY and COUNTRY tables, query the names of all the continents (COUNTRY.Continent) and their respective average city populations (CITY.Population) rounded down to the nearest integer.

Note: CITY.CountryCode and COUNTRY.Code are matching key columns. Do not include continents without cities in your output.

我的解决方案：

SELECT COUNTRY.CONTINENT, FLOOR(AVG(CITY.POPULATION)) FROM COUNTRY INNER JOIN CITY ON COUNTRY.CODE=CITY.COUNTRYCODE

但这似乎不起作用，直到我添加 GROUP BY 语句更新的解决方案：

SELECT COUNTRY.CONTINENT, FLOOR(AVG(CITY.POPULATION)) FROM COUNTRY INNER JOIN CITY ON COUNTRY.CODE=CITY.COUNTRYCODE GROUP BY COUNTRY.CONTINENT

为什么会这样？为什么新INNER JOIN table的平均人口值没有显示出来？我知道这会给我错误的答案，即它会为每个大陆显示相同的平均人口值。但我的疑问是为什么当我不添加 GROUP BY 语句时它不起作用。

抛出的错误：

ERROR 1140 (42000) at line 1: In aggregated query without GROUP BY, expression #1 of SELECT list contains nonaggregated column 'run_y53padyvlle.COUNTRY.continent'; this is incompatible with sql_mode=only_full_group_by

Answer 1

问题不在于 AVG()。你可以这样做：

SELECT FLOOR(AVG(ci.POPULATION))
FROM COUNTRY c INNER JOIN
     CITY ci
     ON c.CODE = ci.COUNTRYCODE;

这一returns行是数据库中所有城市的总体平均人口。

问题是当你这样做的时候：

SELECT c.CONTINENT, FLOOR(AVG(ci.POPULATION)) 
. . .

CONTINENT 部分未聚合。 SQL 引擎需要知道如何处理它。将密钥放入 GROUP BY:

GROUP BY c.CONTINENT

表示您希望 CONTINENT 中的每个值在结果集中占一行。

聚合函数是否需要 GROUP BY？

Is GROUP BY needed for aggregation function?

mysql

sql

group-by

aggregation