SELECT TIMESTAMPDIFF(SECOND,'2018-01-19 14:17:15','2018-01-20 14:17:15');

第二种方法

选择(DATEDIFF(1993-02-20)、1993-02-19)

CURRENT_TIME() --this will return current Date
DATEDIFF('','') --this function will return  DAYS and in 1 day there are 24hh 60mm 60sec

为什么不干脆

从表中选择Sum(Date1 - Date2)

Date1和date2是datetime

SELECT TIMESTAMPDIFF(SECOND,'2018-01-19 14:17:15','2018-01-20 14:17:15');

第二种方法

选择(DATEDIFF(1993-02-20)、1993-02-19)

CURRENT_TIME() --this will return current Date
DATEDIFF('','') --this function will return  DAYS and in 1 day there are 24hh 60mm 60sec

你可以简单地这样做:

SELECT (end_time - start_time) FROM t; -- return in Millisecond
SELECT (end_time - start_time)/1000 FROM t; -- return in Second

如果你在文本字段中存储了一个日期作为字符串，你可以实现这段代码，它将获取过去一周，一个月或一年排序的天数列表:

SELECT * FROM `table` WHERE STR_TO_DATE(mydate, '%d/%m/%Y') < CURDATE() - INTERVAL 30 DAY AND STR_TO_DATE(date, '%d/%m/%Y') > CURDATE() - INTERVAL 60 DAY

//This is for a month

SELECT * FROM `table` WHERE STR_TO_DATE(mydate, '%d/%m/%Y') < CURDATE() - INTERVAL 7 DAY AND STR_TO_DATE(date, '%d/%m/%Y') > CURDATE() - INTERVAL 14 DAY

//This is for a week 

%d%m%Y是日期格式

This query display the record between the days you set there like: Below from last 7 days and Above from last 14 days so it would be your last week record to be display same concept is for month or year. Whatever value you're providing in below date like: below from 7-days so the other value would be its double as 14 days. What we are saying here get all records above from last 14 days and below from last 7 days. This is a week record you can change value to 30-60 days for a month and also for a year.

希望它能帮助到别人。

此代码以yyyy MM dd格式计算两个日期之间的差值。

declare @StartDate datetime 
declare @EndDate datetime

declare @years int
declare @months int 
declare @days int

--NOTE: date of birth must be smaller than As on date, 
--else it could produce wrong results
set @StartDate = '2013-12-30' --birthdate
set @EndDate  = Getdate()            --current datetime

--calculate years
select @years = datediff(year,@StartDate,@EndDate)

--calculate months if it's value is negative then it 
--indicates after __ months; __ years will be complete
--To resolve this, we have taken a flag @MonthOverflow...
declare @monthOverflow int
select @monthOverflow = case when datediff(month,@StartDate,@EndDate) - 
  ( datediff(year,@StartDate,@EndDate) * 12) <0 then -1 else 1 end
--decrease year by 1 if months are Overflowed
select @Years = case when @monthOverflow < 0 then @years-1 else @years end
select @months =  datediff(month,@StartDate,@EndDate) - (@years * 12) 

--as we do for month overflow criteria for days and hours 
--& minutes logic will followed same way
declare @LastdayOfMonth int
select @LastdayOfMonth =  datepart(d,DATEADD
    (s,-1,DATEADD(mm, DATEDIFF(m,0,@EndDate)+1,0)))

select @days = case when @monthOverflow<0 and 
    DAY(@StartDate)> DAY(@EndDate) 
then @LastdayOfMonth + 
  (datepart(d,@EndDate) - datepart(d,@StartDate) ) - 1  
      else datepart(d,@EndDate) - datepart(d,@StartDate) end 


select
 @Months=case when @days < 0 or DAY(@StartDate)> DAY(@EndDate) then @Months-1 else @Months end

Declare @lastdayAsOnDate int;
set @lastdayAsOnDate = datepart(d,DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,@EndDate),0)));
Declare @lastdayBirthdate int;
set @lastdayBirthdate =  datepart(d,DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,@StartDate)+1,0)));

if (@Days < 0) 
(
    select @Days = case when( @lastdayBirthdate > @lastdayAsOnDate) then
        @lastdayBirthdate + @Days
    else
        @lastdayAsOnDate + @Days
    end
)
print  convert(varchar,@years)   + ' year(s),   '  +
       convert(varchar,@months)  + ' month(s),   ' +
       convert(varchar,@days)    + ' day(s)   '