GuilinDev

Lc0197

05 August 2008

197 Rising Temperature

Table: 

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Weather


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+---------------+---------+
| 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’ 

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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:

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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 (20 -> 30).

Schema

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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')

Solution

TO_DAYS()

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# Write your MySQL query statement below
Select w_cur.ID From weather w_cur, weather w_prev 
Where w_cur.temperature > w_prev.temperature And
    TO_DAYS(w_cur.recordDate ) - TO_DAYS(w_prev.recordDate) = 1
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