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中文文档

Description

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 on a certain day.

 

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

Return the result table in any order.

The query result format is in the following example.

 

Example 1:

Input: 
Weather table:
+----+------------+-------------+
| 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  |
+----+
Explanation: 
In 2015-01-02, the temperature was higher than the previous day (10 -> 25).
In 2015-01-04, the temperature was higher than the previous day (20 -> 30).

Solutions

SQL

SELECT w1.Id
FROM
    Weather AS w1,
    Weather AS w2
WHERE
    DATEDIFF(w1.RecordDate, w2.RecordDate) = 1
    AND w1.Temperature > w2.Temperature;
SELECT
	w2.id AS Id
FROM
	weather AS w1
	JOIN weather AS w2 ON DATE_ADD( w1.recordDate, INTERVAL 1 DAY) = w2.recordDate
WHERE
	w1.temperature < w2.temperature