题目描述
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脚本
-- ----------------------------
-- Table structure for `weather`
-- ----------------------------
DROP TABLE IF EXISTS `weather`;
CREATE TABLE `weather` (
`Id` int(11) DEFAULT NULL,
`RecordDate` date DEFAULT NULL,
`Temperature` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
-- ----------------------------
-- Records of weather
-- ----------------------------
INSERT INTO `weather` VALUES ('1','2015-01-01', '10');
INSERT INTO `weather` VALUES ('2','2015-01-02', '25');
INSERT INTO `weather` VALUES ('3','2015-01-03', '20');
INSERT INTO `weather` VALUES ('4','2015-01-04', '30');
本题答案
我们可以使用MySQL的函数DATEDIFF
来计算两个日期的差值,我们的限制条件是温度高且日期差1。
# Write your MySQL query statement below
SELECT a.Id
FROM Weather a INNER JOIN Weather b
WHERE DATEDIFF(a.RecordDate,b.RecordDate) = 1 AND a.Temperature > b.Temperature;