
SQL 架构:
Create table If Not Exists Weather (Id int, RecordDate date, Temperature int)
Truncate table Weather
insert into Weather (Id, RecordDate, Temperature) values ('1', '2015-01-01', '10')
insert into Weather (Id, RecordDate, Temperature) values ('2', '2015-01-02', '25')
insert into Weather (Id, RecordDate, Temperature) values ('3', '2015-01-03', '20')
insert into Weather (Id, RecordDate, Temperature) values ('4', '2015-01-04', '30')
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 |
+----+
MySQL:
# Write your MySQL query statement below
SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND DATEDIFF(w1.RecordDate, w2.RecordDate) = 1;
我的输入
{"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]]}}
我的答案
{"headers":["Id"],"values":[[2],[4]]}




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