Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.

+---------+------------+------------------+| Id(INT) | Date(DATE) | Temperature(INT) |+---------+------------+------------------+|       1 | 2015-01-01 |               10 ||       2 | 2015-01-02 |               25 ||       3 | 2015-01-03 |               20 ||       4 | 2015-01-04 |               30 |+---------+------------+------------------+

For example, return the following Ids for the above Weather table:

+----+| Id |+----+|  2 ||  4 |+----+

要点：

• 比较同一个表中上下两行数据       解法：使用自连接
• 日期函数  datediff(date1,date2) #date1-date2

SELECT Weather.Id as IdFROM WeatherJOIN Weather wON datediff(Weather.Date,w.Date)=1 AND w.Temperature