# Write your MySQL query statement below
1294. 不同国家的天气类型
表:Countries
+---------------+---------+ | Column Name | Type | +---------------+---------+ | country_id | int | | country_name | varchar | +---------------+---------+ country_id 是这张表的主键(具有唯一值的列)。 该表的每行有 country_id 和 country_name 两列。
表:Weather
+---------------+---------+ | Column Name | Type | +---------------+---------+ | country_id | int | | weather_state | varchar | | day | date | +---------------+---------+ (country_id, day) 是该表的复合主键(具有唯一值的列)。 该表的每一行记录了某个国家某一天的天气情况。
编写解决方案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:
weather_state
的平均值小于或等于 15
返回 Cold,weather_state
的平均值大于或等于 25
返回 Hot,以 任意顺序 返回你的查询结果。
返回结果格式如下所示:
示例 1:
输入: 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 | +------------+---------------+------------+ 输出: +--------------+--------------+ | country_name | weather_type | +--------------+--------------+ | USA | Cold | | Austraila | Cold | | Peru | Hot | | China | Warm | | Morocco | Hot | +--------------+--------------+ 解释: 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。
原站题解
mysql 解法, 执行用时: 485 ms, 内存消耗: 0 B, 提交时间: 2023-10-15 16:36:06
# Write your MySQL query statement below SELECT country_name AS 'country_name', CASE WHEN AVG(w1.weather_state) <= 15 THEN 'Cold' WHEN AVG(w1.weather_state) >= 25 THEN 'Hot' ELSE 'Warm' END AS 'weather_type' FROM Countries AS c1 INNER JOIN Weather AS w1 ON c1.country_id = w1.country_id WHERE DATE_FORMAT(w1.day, '%Y-%m') = '2019-11' GROUP BY c1.country_id ;
mysql 解法, 执行用时: 380 ms, 内存消耗: 0 B, 提交时间: 2023-10-15 16:35:55
# Write your MySQL query statement below SELECT country_name AS 'country_name', CASE WHEN AVG(w1.weather_state) <= 15 THEN 'Cold' WHEN AVG(w1.weather_state) >= 25 THEN 'Hot' ELSE 'Warm' END AS 'weather_type' FROM Countries AS c1 INNER JOIN Weather AS w1 ON c1.country_id = w1.country_id WHERE YEAR(w1.day) = 2019 AND MONTH(w1.day) = 11 GROUP BY c1.country_id ;
mysql 解法, 执行用时: 490 ms, 内存消耗: 0 B, 提交时间: 2023-10-15 16:35:41
# Write your MySQL query statement below SELECT country_name AS 'country_name', CASE WHEN AVG(w1.weather_state) <= 15 THEN 'Cold' WHEN AVG(w1.weather_state) >= 25 THEN 'Hot' ELSE 'Warm' END AS 'weather_type' FROM Countries AS c1 INNER JOIN Weather AS w1 ON c1.country_id = w1.country_id WHERE w1.day BETWEEN '2019-11-01' AND '2019-11-30' GROUP BY c1.country_id ;