如何计算两个日期之间的差异，格式为YYYY-MM-DD hh: mm: ss，并以秒或毫秒为单位得到结果？
答案 0 :(得分：279)
SELECT TIMEDIFF('2007-12-31 10:02:00','2007-12-30 12:01:01');
-- result: 22:00:59, the difference in HH:MM:SS format
SELECT TIMESTAMPDIFF(SECOND,'2007-12-30 12:01:01','2007-12-31 10:02:00');
-- result: 79259 the difference in seconds
因此，您可以将TIMESTAMPDIFF用于您的目的。
答案 1 :(得分：34)
如果您正在使用DATE列（或者可以将它们转换为日期列），请尝试DATEDIFF（）然后乘以24小时，60分钟，60秒（因为DATEDIFF以天为单位返回diff）。来自MySQL：
http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html
例如：
mysql> SELECT DATEDIFF('2007-12-31 23:59:59','2007-12-30 00:00:00') * 24*60*60
答案 2 :(得分：26)
使用DATEDIFF
获取日期差异SELECT DATEDIFF('2010-10-08 18:23:13', '2010-09-21 21:40:36') AS days;
+------+
| days |
+------+
| 17 |
+------+
或
答案 3 :(得分：8)
SELECT TIMESTAMPDIFF(HOUR,NOW(),'2013-05-15 10:23:23')
calculates difference in hour.(for days--> you have to define day replacing hour
SELECT DATEDIFF('2012-2-2','2012-2-1')
SELECT TO_DAYS ('2012-2-2')-TO_DAYS('2012-2-1')
答案 4 :(得分：4)
select
unix_timestamp('2007-12-30 00:00:00') -
unix_timestamp('2007-11-30 00:00:00');
答案 5 :(得分：1)
SELECT TIMESTAMPDIFF(SECOND,'2018-01-19 14:17:15','2018-01-20 14:17:15');
第二种方法
SELECT ( DATEDIFF('1993-02-20','1993-02-19')*( 24*60*60) )AS 'seccond';
CURRENT_TIME() --this will return current Date
DATEDIFF('','') --this function will return DAYS and in 1 day there are 24hh 60mm 60sec
答案 6 :(得分：0)
或者，您可以使用TIMEDIFF功能
mysql> SELECT TIMEDIFF('2000:01:01 00:00:00', '2000:01:01 00:00:00.000001');
'-00:00:00.000001'
mysql> SELECT TIMEDIFF('2008-12-31 23:59:59.000001' , '2008-12-30 01:01:01.000002');
'46:58:57.999999'
答案 7 :(得分：0)
此功能获取两个日期之间的差异，并以日期格式yyyy-mm-dd显示。您只需要执行下面的代码然后使用该功能。执行后你可以像这样使用它
SELECT datedifference(date1, date2)
FROM ....
.
.
.
.
DELIMITER $$
CREATE FUNCTION datedifference(date1 DATE, date2 DATE) RETURNS DATE
NO SQL
BEGIN
DECLARE dif DATE;
IF DATEDIFF(date1, DATE(CONCAT(YEAR(date1),'-', MONTH(date1), '-', DAY(date2)))) < 0 THEN
SET dif=DATE_FORMAT(
CONCAT(
PERIOD_DIFF(date_format(date1, '%y%m'),date_format(date2, '%y%m'))DIV 12 ,
'-',
PERIOD_DIFF(date_format(date1, '%y%m'),date_format(date2, '%y%m'))% 12 ,
'-',
DATEDIFF(date1, DATE(CONCAT(YEAR(date1),'-', MONTH(DATE_SUB(date1, INTERVAL 1 MONTH)), '-', DAY(date2))))),
'%Y-%m-%d');
ELSEIF DATEDIFF(date1, DATE(CONCAT(YEAR(date1),'-', MONTH(date1), '-', DAY(date2)))) < DAY(LAST_DAY(DATE_SUB(date1, INTERVAL 1 MONTH))) THEN
SET dif=DATE_FORMAT(
CONCAT(
PERIOD_DIFF(date_format(date1, '%y%m'),date_format(date2, '%y%m'))DIV 12 ,
'-',
PERIOD_DIFF(date_format(date1, '%y%m'),date_format(date2, '%y%m'))% 12 ,
'-',
DATEDIFF(date1, DATE(CONCAT(YEAR(date1),'-', MONTH(date1), '-', DAY(date2))))),
'%Y-%m-%d');
ELSE
SET dif=DATE_FORMAT(
CONCAT(
PERIOD_DIFF(date_format(date1, '%y%m'),date_format(date2, '%y%m'))DIV 12 ,
'-',
PERIOD_DIFF(date_format(date1, '%y%m'),date_format(date2, '%y%m'))% 12 ,
'-',
DATEDIFF(date1, DATE(CONCAT(YEAR(date1),'-', MONTH(date1), '-', DAY(date2))))),
'%Y-%m-%d');
END IF;
RETURN dif;
END $$
DELIMITER;
答案 8 :(得分：0)
select TO_CHAR(TRUNC(SYSDATE)+(to_date( '31-MAY-2012 12:25', 'DD-MON-YYYY HH24:MI')
- to_date( '31-MAY-2012 10:37', 'DD-MON-YYYY HH24:MI')),
'HH24:MI:SS') from dual
- 结果：01：48：00
好吧，这不是OP所要求的，但这是我想做的事情： - ）
答案 9 :(得分：0)
此代码以yyyy MM dd格式计算两个日期之间的差异。
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) '
答案 10 :(得分：0)
如果您将日期存储在文本字段中作为字符串，则可以实现此代码，它将获取一周，一个月或一年排序的过去天数列表：
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是您的日期格式
此查询显示您在此处设置的日期之间的记录，例如：过去7天以下和过去14天以上的记录，因此您的上周记录显示相同的概念是月份或年份。无论您在下面的日期提供的价值如下：7天以下，其他价值将是14天的两倍。我们在这里说的是从过去7天以来的所有记录中获取的所有记录。这是一周记录，您可以将值更改为30-60天，一个月以及一年。
谢谢你希望它会帮助别人。
答案 11 :(得分：-1)
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
答案 12 :(得分：-1)
为什么不
从表
中选择Sum（Date1 - Date2）date1和date2是日期时间