the leetcode question 197.Rising Temperature
Given a Weather table, write a SQL query to find all dates’ Ids with higher temperature compared to its previous (yesterday’s) dates.
x
+---------+------------------+------------------+
| 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 |
+----+
i didn’t know the function DATEDIFF()
,so i wrote my sql solution:
select w1.Id
from Weather w1,Weather w2
where w1.RecordDate - w2.RecordDate = 1
and w1.Temperature > w2.Temperature
and i went through the testcase but got a wrong submit,the right solution is use funtion DATEDIFF()
select w1.Id
from Weather w1,Weather w2
where DATEDIFF(w1.RecordDate,w2.RecordDate)=1
and w1.Temperature > w2.Temperature
so my question is what’s the difference between
DATEDIFF(w1.RecordDate,w2.RecordDate)=1
and w1.RecordDate - w2.RecordDate = 1
thank you for your help
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Answer
If the datatype of RecordDate
is DATETIME
rather than DATE
, subtracting them returns a large value that contains the difference between the times as well as the dates. E.g.
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 |
+-----------------+
But if they’re DATE
then subtraction should be the same as DATEDIFF()
:
mysql> select cast('2019-03-21' as date) - cast('2019-03-11' as date) as difference;
+------------+
| difference |
+------------+
| 10 |
+------------+