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 |
+----+
利用 datediff 函数返回两个日期相差的天数。
# Write your MySQL query statement below
select a.Id
from Weather a
inner join Weather b
on a.Temperature > b.Temperature and datediff(a.RecordDate, b.RecordDate) = 1;
{"headers": {"Weather": ["Id", "RecordDate", "Temperature"]}, "rows": {"Weather": [[1, "2015-01-01", 10], [2, "2015-01-02", 25], [3, "2015-01-03", 20], [4, "2015-01-04", 30]]}}
{"headers":["Id"],"values":[[2],[4]]}