sql日期时间相减语句
本款教程利用了datediff函数,来对数据库教程的日期进行相减查询哦,下面便写了n种关于mssql 日期相减的方法。
sql中两个日期相减
1、相差天数
select trunc(sysdate,"yyyy")-to_date("2009-01-04","yyyy-mm-dd") from dual;
2、相差月数
select months_between(trunc(sysdate,"mm"),to_date("2009-01","yyyy-mm")) from dual;
3、相差年数
select trunc((months_between(trunc(sysdate,"dd"),to_date("2009-01-01","yyyy-mm-dd")))/12) from dual;
4、select datediff( day, "2008.08.25", "2008.09.01" )
5、select datediff( second, 2009-8-25 12:15:12", 2009-9-1 7:18:20") --返回相差秒数
6、
select datediff( minute, 2009-8-25 12:15:12", 2009-9-1 7:18:20") --返回相差分钟数
7、
select datediff( hour, 2009-8-25 12:15:12", 2009-9-1 7:18:20") --返回相差小时数
问题三:select datediff( day, 2009-8-25 12:15:12", 2009-9-1 7:18:20")
实例二
use pubs
select distinct datediff(day, "2009-3-12", "2009-3-15") as difday
from titles
结果:3
declare @dt1 as datetime, @dt2 as datetime;
select @dt1 = "2008-8-4 9:36:41", @dt2 = "2008-8-2 9:33:39";
declare @days as int, @hours as int, @minutes as int, @seconds as int;
set @seconds = datediff( second, @dt2, @dt1);
set @days = @seconds / (24 * 60 * 60)
set @seconds = @seconds - @days * 24 * 60 * 60
set @hours = @seconds / (60 * 60);
set @seconds = @seconds - @hours * 60 * 60
set @minutes = @seconds / 60;
set @seconds = @seconds - @minutes * 60;
select convert(varchar(10), @days ) + "天" + convert(varchar(10), @hours ) + "小时" + convert(varchar(10), @minutes ) + "分" + convert(varchar(10), @seconds ) + "秒";
下面来看个实例
我有一个表,其中有四个字段:开始天数,开始时间,到达天数,到达时间(这四个字段都是varchar类型)
例如:某一条记录: 1 16:00 2 12:20
我的目的就是用 select(到达天数+到达时间)-(开始天数+开始时间) as 花费时间 from table
例如上条记录得到的就是 (2*24:00+12:20)-(24:00+16:00)=20:00
这样的sql语句该怎么写???
declare @t table
(
beginday int,
begintime varchar(20),
endday int,
endtime varchar(20)
)
insert @t select 1,"16:00",2,"12:20"
union all select 1,"3:00",3,"19:10"
select
date=rtrim(date/60)+":"+rtrim(date%60)
from
(select date=datediff(mi,1,dateadd(d,endday-beginday,beginday)-begintime+endtime)from @t )t
date
-------------------------
20:20
64:10
方法二
declare @t table(开始天数 varchar(10),开始时间 varchar(10),到达天数 varchar(10),到达时间 varchar(10))
insert @t select "1", "16:00","2","12:20"
--如果开始天数,到达天数大于31
select 到达天数 * 24 + datepart(hh,到达时间) - 开始天数 * 24 - datepart(hh,开始时间)
from @t
更多推荐
数据库中的日期相减_sql日期相减得到天数【sql日期时间相减语句】
发布评论