mysql - DATEDIFF 和自己做减法有什么区别

Question

leetcode问题197.Rising Temperature

---

给定一个天气表，编写一个 SQL 查询来查找与之前(昨天)日期相比温度较高的所有日期 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 |
+---------+------------------+------------------+
For example, return the following Ids for the above Weather table:

+----+
| Id |
+----+
|  2 |
|  4 |
+----+

---

我不知道函数DATEDIFF()，所以我写了我的sql解决方案:

select w1.Id
from Weather w1,Weather w2
where w1.RecordDate - w2.RecordDate = 1
and w1.Temperature > w2.Temperature

我完成了测试用例，但提交错误，正确的解决方案是使用函数 DATEDIFF()

select w1.Id
from Weather w1,Weather w2
where DATEDIFF(w1.RecordDate,w2.RecordDate)=1
and w1.Temperature > w2.Temperature

所以我的问题是两者之间有什么区别 DATEDIFF(w1.RecordDate,w2.RecordDate)=1和w1.RecordDate - w2.RecordDate = 1

感谢您的帮助

Answer 1

如果 RecordDate 的数据类型是 DATETIME 而不是 DATE，则减去它们会返回一个大值，其中也包含时间之间的差异作为日期。例如

mysql> select cast('2019-03-21 10:20:30' as datetime) - cast('2019-03-11 9:15:20' as datetime) as difference;
+-----------------+
| difference      |
+-----------------+
| 10010510.000000 |
+-----------------+

但如果它们是DATE，那么减法应该与DATEDIFF()相同:

mysql> select cast('2019-03-21' as date) - cast('2019-03-11' as date) as difference;
+------------+
| difference |
+------------+
|         10 |
+------------+