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 |
+----+
解法：
SELECT w1.Id FROM Weather w1
INNER JOIN Weather w2 ON
w1.Temperature > w2.Temperature AND DATEDIFF(w1.Date, w2.Date) = 1