题目
国家表：Countries
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| country_id | int |
| country_name | varchar |
+---------------+---------+
country_id 是这张表的主键。
该表的每行有 country_id 和 country_name 两列。
天气表：Weather
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| country_id | int |
| weather_state | varchar |
| day | date |
+---------------+---------+
(country_id, day) 是该表的复合主键。
该表的每一行记录了某个国家某一天的天气情况。
写一段 SQL 来找到表中每个国家在 2019 年 11 月的天气类型。
天气类型的定义如下：当 weather_state 的平均值小于或等于15返回 Cold，当 weather_state 的平均值大于或等于 25 返回 Hot，否则返回 Warm。
你可以以任意顺序返回你的查询结果。
查询结果格式如下所示：
Countries table:
+------------+--------------+
| country_id | country_name |
+------------+--------------+
| 2 | USA |
| 3 | Australia |
| 7 | Peru |
| 5 | China |
| 8 | Morocco |
| 9 | Spain |
+------------+--------------+
Weather table:
+------------+---------------+------------+
| country_id | weather_state | day |
+------------+---------------+------------+
| 2 | 15 | 2019-11-01 |
| 2 | 12 | 2019-10-28 |
| 2 | 12 | 2019-10-27 |
| 3 | -2 | 2019-11-10 |
| 3 | 0 | 2019-11-11 |
| 3 | 3 | 2019-11-12 |
| 5 | 16 | 2019-11-07 |
| 5 | 18 | 2019-11-09 |
| 5 | 21 | 2019-11-23 |
| 7 | 25 | 2019-11-28 |
| 7 | 22 | 2019-12-01 |
| 7 | 20 | 2019-12-02 |
| 8 | 25 | 2019-11-05 |
| 8 | 27 | 2019-11-15 |
| 8 | 31 | 2019-11-25 |
| 9 | 7 | 2019-10-23 |
| 9 | 3 | 2019-12-23 |
+------------+---------------+------------+
Result table:
+--------------+--------------+
| country_name | weather_type |
+--------------+--------------+
| USA | Cold |
| Austraila | Cold |
| Peru | Hot |
| China | Warm |
| Morocco | Hot |
+--------------+--------------+
USA 11 月的平均 weather_state 为 (15) / 1 = 15 所以天气类型为 Cold。
Australia 11 月的平均 weather_state 为 (-2 + 0 + 3) / 3 = 0.333 所以天气类型为 Cold。
Peru 11 月的平均 weather_state 为 (25) / 1 = 25 所以天气类型为 Hot。
China 11 月的平均 weather_state 为 (16 + 18 + 21) / 3 = 18.333 所以天气类型为 Warm。
Morocco 11 月的平均 weather_state 为 (25 + 27 + 31) / 3 = 27.667 所以天气类型为 Hot。
我们并不知道 Spain 在 11 月的 weather_state 情况所以无需将他包含在结果中。
解答
选出2019年11月的天气情况
select *
from Weather as W
where year(W.day) = 2019 and month(W.year) = 11
两表连接
select tmp.country_name,
if(avg(tmp.weather_state)<=15, 'Cold',if(avg(tmp.weather_state)>=25, 'Hot', 'Warm'))
from Countries as C
join (select *
from Weather as W
where year(W.day) = 2019 and month(W.year) = 11) as tmp
on C.country_id = tmp. country_id
group by tmp.country_name;
别的解答
思路都是类似的 只是先筛选 先连接的区别
select country_name,
case when avg(weather_state)<= 15 then 'Cold'
when avg(weather_state)>= 25 then 'Hot'
else 'Warm'
end as weather_type
from countries c
join weather w
on c.country_id=w.country_id
where day like'2019-11%'
group by country_name;
select c.country_name,
if(avg_wea<= 15,'Cold',if(avg_wea<25,'Warm','Hot')) as weather_type
from
(select country_id,avg(weather_state) as avg_wea from Weather
where day between '2019-11-01' and '2019-11-30'
group by country_id) as w
left join Countries c
on w.country_id = c.country_id;