尝试从两个不同的表中加入列

Question

我有一张表（city_data），其中包含“year”、“city_name”和“avg_temp”列。 另一个表是 global_data 并具有列“avg_temp”和“year”。 我正在尝试创建一个具有伦敦平均温度的表格，即多年来的全球平均温度。 我的 SQL 代码如下：

SELECT city_data.avg_temp, city_data.year
FROM city_data
JOIN global_data
ON city_data.year = global_data.avg_temp
WHERE city= 'Paris'

我没有收到错误，但无法获得预期的结果。 我想我无法理解我需要在 ON 子句中添加什么。

Answer 1

FROM city_data
JOIN global_data
ON city_data.year = global_data.year

-- Join 中应该是global_data.year

WHERE city= 'Paris'

Answer 2

这完全基于部分理解

SELECT city_name, AVG(city_data.avg_temp) as global_avg
    FROM
    city_data
    JOIN global_data
    ON city_data.year = global_data.year
    WHERE city_name = 'Paris'
    GROUP BY city_data.city_name

Answer 3

由于year是唯一的共同属性，因此使用它进行连接。 然后获取城市平均表，因为它在city_data中很容易获得。 最后做global_data的 avg ，因为你想跨年获取它。

SELECT `city_name`, `city_data`.`avg_temp` AS city_avg_temp, AVG(`global_data`.`avg_temp`) AS global_avg_temp
    FROM `city_data`
    JOIN `global_data` 
        ON `city_data`.`year` = `global_data`.`year`
    WHERE `city_name` = 'London'
    GROUP BY `city_data`.`city_name`

Answer 4

我想表global_data每年有一行，而city_data每年有一行，城市。 这意味着您可以简单地在他们共同的年份加入两者：

select year, g.avg_temp as global_average, c.avg_temp as paris_average
from global_data g
join city_data c using (year)
where c.city = 'Paris'
order by year;

如果你不喜欢加入两者然后才提到巴黎，你也可以把它变成：

select year, g.avg_temp as global_average, p.avg_temp as paris_average
from global_data g
join (select * from city_data where city = 'Paris') p using (year)
order by year;

从 MySQL 8 开始，您甚至可以使用WITH子句来“准备”巴黎:-)

with paris as (select * from city_data where city = 'Paris')
select year, g.avg_temp as global_average, paris.avg_temp as paris_average
from global_data g
join paris using (year)
order by year;

或将巴黎视为加入标准：

select g.year, g.avg_temp as global_average, p.avg_temp as paris_average
from global_data g
join city_data p on p.year = g.year and p.city = 'Paris'
order by g.year;

如您所见，编写相同查询的方法有很多种，并且 DBMS 可能会为所有人提供相同的执行计划。 因此，只需选择您最喜欢的查询。