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 | +----+
查找与昨天的日期相比更高的日期的id。
mysql(1868ms):
SELECT w1.Id FROM Weather w1 , Weather w2 WHERE w1.Temperature > w2.Temperature AND TO_DAYS(w1.Date) - TO_DAYS(w2.Date) = 1 ;