SQLSERVER时间函数
2011-07-12 13:49 bingcaihuang 阅读(644) 评论(0) 编辑 收藏 举报一、sql server日期时间函数Sql Server中的日期与时间函数
1. 当前系统日期、时间
select getdate()
2. dateadd 在向指定日期加上一段时间的基础上,返回新的 datetime 值
例如:向日期加上2天
select dateadd(day,2,'2004-10-15') --返回:2004-10-17 00:00:00.000
3. datediff 返回跨两个指定日期的日期和时间边界数。
select datediff(day,'2004-09-01','2004-09-18') --返回:17
4. datepart 返回代表指定日期的指定日期部分的整数。
select DATEPART(month, '2004-10-15') --返回 10
5. datename 返回代表指定日期的指定日期部分的字符串
select datename(weekday, '2004-10-15') --返回:星期五
6. day(), month(),year() --可以与datepart对照一下
select 当前日期=convert(varchar(10),getdate(),120)
,当前时间=convert(varchar(8),getdate(),114)
select datename(dw,'2004-10-15')
select 本年第多少周=datename(week,'2004-10-15')
,今天是周几=datename(weekday,'2004-10-15')
select CONVERT(varchar, getdate(), 120 )
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1.GetDate() 用于sql server :select GetDate()
2.DateDiff('s','2005-07-20','2005-7-25 22:56:32')返回值为 514592 秒
DateDiff('d','2005-07-20','2005-7-25 22:56:32')返回值为 5 天
3.DatePart('w','2005-7-25 22:56:32')返回值为 2 即星期一(周日为1,周六为7)
DatePart('d','2005-7-25 22:56:32')返回值为 25即25号
DatePart('y','2005-7-25 22:56:32')返回值为 206即这一年中第206天
DatePart('yyyy','2005-7-25 22:56:32')返回值为 2005即2005年
附图
函数 | 参数/功能 |
GetDate( ) | 返回系统目前的日期与时间 |
DateDiff (interval,date1,date2) | 以interval 指定的方式,返回date2 与date1两个日期之间的差值 date2-date1 |
DateAdd (interval,number,date) | 以interval指定的方式,加上number之后的日期 |
DatePart (interval,date) | 返回日期date中,interval指定部分所对应的整数值 |
DateName (interval,date) | 返回日期date中,interval指定部分所对应的字符串名称 |
参数 interval的设定值如下:
值 | 缩 写(Sql Server) | Access 和 ASP | 说明 |
Year | Yy | yyyy | 年 1753 ~ 9999 |
Quarter | q | 季 1 ~ 4 | |
Month | Mm | m | 月1 ~ 12 |
Day of year | Dy | y | 一年的日数,一年中的第几日 1-366 |
Day | Dd | d | 日,1-31 |
Weekday | Dw | w | 一周的日数,一周中的第几日 1-7 |
Week | Wk | ww | 周,一年中的第几周 0 ~ 51 |
Hour | Hh | h | 时0 ~ 23 |
Minute | Mi | n | 分钟0 ~ 59 |
Second | Ss | s | 秒 0 ~ 59 |
Millisecond | Ms | - | 毫秒 0 ~ 999 |
转自:http://www.cnblogs.com/vic_lu/archive/2010/12/17/1909135.html