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by Sumit Chourasia | Oct 16, 2020 | Category :coding | Tags : algorithm data-structure easy leetcode mysql sql

Rising Temperature - SQL - Easy - LeetCode

Rising Temperature - SQL - Easy - LeetCode

Table: Weather

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| recordDate    | date    |
| temperature   | int     |
+---------------+---------+
id is the primary key for this table.
This table contains information about the temperature in a certain day.
 

Write an SQL query to find all dates' id with higher temperature compared to its previous dates (yesterday).

Return the result table in any order.

The query result format is in the following example:

Weather
+----+------------+-------------+
| id | recordDate | Temperature |
+----+------------+-------------+
| 1  | 2015-01-01 | 10          |
| 2  | 2015-01-02 | 25          |
| 3  | 2015-01-03 | 20          |
| 4  | 2015-01-04 | 30          |
+----+------------+-------------+

Result table:
+----+
| id |
+----+
| 2  |
| 4  |
+----+
In 2015-01-02, temperature was higher than the previous day (10 -> 25).
In 2015-01-04, temperature was higher than the previous day (30 -> 20).

# Write your MySQL query statement below
select w1.id from Weather w1, Weather w2 Where w1.Temperature > w2.Temperature And TO_DAYS(w1.recordDate)-TO_DAYS(w2.recordDate)=1;

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