给定一个 Weather
表,编写一个 SQL 查询,来查找与之前(昨天的)日期相比温度更高的所有日期的 Id。
+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
| 1 | 2015-01-01 | 10 |
| 2 | 2015-01-02 | 25 |
| 3 | 2015-01-03 | 20 |
| 4 | 2015-01-04 | 30 |
+---------+------------------+------------------+
例如,根据上述给定的 Weather
表格,返回如下 Id:
+----+
| Id |
+----+
| 2 |
| 4 |
+----+
思路:SELF JOIN 表,然后设置条件查询,语法:
TO_DAYS(wt1.DATE) return the number of days between from year 0 to date DATE
SELECT w1.ID AS ID
FROM Weather AS w1, Weather AS w2
WHERE (w1.Temperature > w2.Temperature)
AND (TO_DAYS(w1.RecordDate)-TO_DAYS(w2.RecordDate)=1)