Leetcode 197 Rising temperature (July 16, 2022)

SQL framework ：

Create table If Not Exists Weather (id int, recordDate date, temperature int)
Truncate table Weather
insert into Weather (id, recordDate, temperature) values ('1', '2015-01-01', '10')
insert into Weather (id, recordDate, temperature) values ('2', '2015-01-02', '25')
insert into Weather (id, recordDate, temperature) values ('3', '2015-01-03', '20')
insert into Weather (id, recordDate, temperature) values ('4', '2015-01-04', '30')

surface ： Weather

±--------------±--------+
| Column Name | Type |
±--------------±--------+
| id | int |
| recordDate | date |
| temperature | int |
±--------------±--------+
id It's the primary key of this table
The table contains temperature information for a specific date

Write a SQL Inquire about , To find out （ Yesterday's ） Of all dates with a higher temperature id .

Return results Order is not required .

The query result format is as follows .

Example 1：

Input ：
Weather surface ：
±—±-----------±------------+
| id | recordDate | Temperature |
±—±-----------±------------+
| 1 | 2015-01-01 | 10 |
| 2 | 2015-01-02 | 25 |
| 3 | 2015-01-03 | 20 |
| 4 | 2015-01-04 | 30 |
±—±-----------±------------+
Output ：
±—+
| id |
±—+
| 2 |
| 4 |
±—+
explain ：
2015-01-02 The temperature is higher than the day before （10 -> 25）
2015-01-04 The temperature is higher than the day before （20 -> 30）

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/rising-temperature

Method 1 ： Use DATEDIFF()

MySQL Submission ：

# Write your MySQL query statement below
SELECT w2.Id
FROM Weather w1, Weather w2
WHERE DATEDIFF(w2.RecordDate, w1.RecordDate) = 1
AND w1.Temperature < w2.Temperature