leetcode原文引用: Given aWeathertable, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates. +---+---+---+ | Id(INT) | Date(DATE) | Temperature(INT) | +---+---+---+ | 1 | 2015-01-01 | 10 | | 2 | 2015-01-0...
解法一: SELECTw1.IdFROMWeather w1, Weather w2WHEREw1.Temperature>w2.TemperatureANDDATEDIFF(w1.Date, w2.Date)=1; 下面这种解法我们使用了MySQL的TO_DAYS函数,用来将日期换算成天数,其余跟上面相同: 解法二: SELECTw1.IdFROMWeather w1, Weather w2WHEREw1.Temperature>w2.TemperatureANDTO_DAYS(w1.Date)=...
https://leetcode.com/problems/rising-temperature/ 之前交了几次都没加TO_DAYS函数,我错了,我不该懒得装MySQL。 selectA.IdfromWeather AwhereA.Temperature>(selectB.TemperaturefromWeather BwhereTO_DAYS(A.Date)-1=TO_DAYS(B.Date))
用sql的DateDiff函数判断。 不知道为何用的和网上的结论相反 DateDiff(b, a) = 1表示b是a的下一天,才能ac # Write your MySQL query statement belowselectb.IdasIdfromWeatherjoin(Weatherasb)onDATEDIFF(b.Date, weather.Date)=1andb.Temperature>weather.Temperature;...
SQL架构 1Create table If Not Exists Weather (Idint, RecordDate date, Temperatureint)2Truncate table Weather3insert into Weather (Id, RecordDate, Temperature) values ('1','2015-01-01','10')4insert into Weather (Id, RecordDate, Temperature) values ('2','2015-01-02','25')5insert into...
Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates. +---+---+---+ | Id(INT) | RecordDate(DATE) | Temperature(INT) | +---+---+---+ | 1 | 2015-01-01 | 10 | |...
Given aWeathertable, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates. 给了一个温度表,找出温度比前一天高的日期id +---+---+---+ | Id(INT) | Date(DATE) | Temperature(INT)...