MySQL中两个日期之间的差异

这段代码计算两个日期之间的yyyy MM dd格式的差异。

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)   '   

如果您在文本字段中存储了日期字符串，您可以实现此代码，它将获取过去一周、一个月或一年的天数列表，并进行排序：

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 是您的日期格式

此查询显示在您设置的日期之间的记录，例如：从过去的7天开始以下和从过去的14天开始以上，因此它将显示您上周的记录，同样的概念也适用于月份或年份。无论您在下面的日期中提供什么值，例如：从7天前开始以下，那么另一个值将是其两倍，即14天。我们在这里说的是获取所有在过去14天内以上并在过去7天内以下的记录。这是一周的记录，您可以将值更改为30-60天以获得一个月的记录，也可以获得一年的记录。

谢谢，希望能帮助到某些人。

你只需要这样做：

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

为什么不直接这样做呢：

从表中选择 Sum(Date1 - Date2)

其中 date1 和 date2 是 datetime 类型