T-SQL: 17 个与日期时间相关的自定义函数(UDF),周日作为周的最后一天,均不受 @@DateFirst、语言版本影响!
CSDN 的 Blog 太滥了!无时不刻地在坏!
开始抢救性搬家 ... ... 到这里重建家园
/*
T-SQL: 17 个与日期时间相关的自定义函数(UDF),周日作为周的最后一天,均不受 @@DateFirst、语言版本影响
都是从老文章里收集或提炼出来的!
提示:
(@@Datefirst + datepart(weekday,@Date)) % 7 判断周几是最保险的! 与 @@DateFirst 无关,与语言版本无关
@@DateFirst 可能会导致 datepart(weekday,@Date) 不一样!
无论 @@DateFirst 等于几,无论是什么语言版本的 SQL Server 下面永远恒成立!
(@@Datefirst + datepart(weekday,@Date))%7 : 2、3、4、5、6、0、1 分别代表 周一 到 周日
-- */
create function udf_GetAge(@StartDate datetime,@EndDate datetime)
returns integer
-- 返回精确年龄 select dbo.udf_GetAge('1949-10-01',getdate())
begin
return datediff(year,@StartDate,@EndDate)
- case when datediff(day,dateadd(year,datediff(year,@StartDate,@EndDate),@StartDate),@EndDate) >= 0
then 0
else
1
end
end
go
create function udf_DaysOfYearByDate(@Date datetime)
returns integer
-- 返回年的天数 可判断 平(365)、润(366) 年
begin
return datediff(day,dateadd(year,datediff(year,0,@Date),0),dateadd(year,datediff(year,0,@Date) + 1,0))
end
go
create function udf_DaysOfYear(@Year integer)
returns integer
-- 返回年的天数 可判断 平(365)、润(366) 年
begin
return datediff(day,dateadd(year,@year - year(0),0),dateadd(year,@year - year(0) + 1,0))
end
go
create function udf_HalfDay(@Date datetime)
returns datetime
-- 返回 @Date 是 上午 返回 @Date 的零点,@Date 是 下午 返回 @Date 的十二点
begin
return case when datepart(hour,@Date) < 12
then dateadd(day,datediff(day,0,@Date),0) --上午归到 零点
else
dateadd(hour,12,dateadd(day,datediff(day,0,@Date),0)) --下午归到 十二点
end
end
go
create function udf_WeekDiff(@StartDate datetime,@EndDate datetime)
returns integer
-- 返回 [@StartDate , @EndDate] 之间周数 周日是当周的最后一天
begin
return datediff(week,@StartDate,@EndDate) -- + 1
+ case when (@@Datefirst + datepart(weekday,@StartDate)) % 7 = 1
then 1
else
0
end
- case when (@@Datefirst + datepart(weekday,@EndDate)) % 7 = 1
then 1
else 0
end
end
go
create function udf_WeekOfMonth(@Date datetime)
-- 返回 @Date 是所在月的第几周 周日是当周的最后一天
returns integer
begin
return datediff(week
,case when (@@Datefirst + datepart(weekday,dateadd(month,datediff(month,0,@Date),0))) % 7 = 1
then dateadd(month,datediff(month,0,@Date),0) - 1
else
dateadd(month,datediff(month,0,@Date),0)
end
,case when (@@Datefirst + datepart(weekday,@Date)) % 7 = 1
then @Date-1
else @Date
end
) + 1
end
go
create function udf_WeekOfQuarter(@Date datetime)
-- 返回 @Date 是所在季度的第几周 周日是当周的最后一天
returns int
begin
return datediff(week
,case when (@@Datefirst + datepart(weekday,dateadd(Quarter,datediff(Quarter,0,@Date),0))) % 7 = 1
then dateadd(Quarter,datediff(Quarter,0,@Date),0) - 1
else
dateadd(Quarter,datediff(Quarter,0,@Date),0)
end
,case when (@@Datefirst + datepart(weekday,@Date)) % 7 = 1
then @Date - 1
else
@Date
end
) + 1
end
go
create function udf_WeekOfYear(@Date datetime)
-- 返回 @Date 是所在年的第几周 周日是当周的最后一天
returns int
begin
return datediff(week
,case when (@@Datefirst + datepart(weekday,dateadd(day,0,datediff(day,0,dateadd(year,datediff(year,0,@Date),0))))) % 7 = 1
then dateadd(day,-1,dateadd(day,0,datediff(day,0,dateadd(year,datediff(year,0,@Date),0))))
else
dateadd(day,0,datediff(day,0,dateadd(year,datediff(year,0,@Date),0))) --date 所在年的第一天 即: 一月一号
end
,case when (@@Datefirst + datepart(weekday,@Date)) % 7 = 1
then dateadd(day,-1,@Date)
else
@Date
end
) + 1
end
go
create function udf_WeekDay(@ int,@Date datetime)
returns datetime
-- 返回 @Date 所在周的其他天 周一 到 周日 也就是映射到 所在周的其他天 周日是当周的最后一天
begin
/*
--周日算作(上一)周的最后一天
当 @ <= 1 代表将 @Date 映射到 所在周的星期一
当 @ = 2 代表将 @Date 映射到 所在周的星期二
当 @ = 3 代表将 @Date 映射到 所在周的星期三
当 @ = 4 代表将 @Date 映射到 所在周的星期四
当 @ = 5 代表将 @Date 映射到 所在周的星期五
当 @ = 6 代表将 @Date 映射到 所在周的星期六
当 @ >= 7 代表将 @Date 映射到 所在周的星期日
可用于按周汇总 Group by,均支持跨年跨月数据
*/
return dateadd(day
,case when (@@Datefirst + datepart(weekday,@Date)) % 7 = 0 --周六
then case when @ between 1 and 6
then @ - 6
else
1
end
when (@@Datefirst + datepart(weekday,@Date)) % 7 = 1 --周日(七)
then case when @ between 1 and 6
then @ - 7
else
0
end
when (@@Datefirst + datepart(weekday,@Date)) % 7 between 2 and 6 --周一至周五
then case when @ between 1 and 6
then @ + 1 - (@@Datefirst + datepart(weekday,@Date)) % 7
else
8 - (@@Datefirst + datepart(weekday,@Date)) % 7
end
end
,@Date)
end
go
create function udf_WeekdayDiff(@Weekday integer,@StartDate datetime,@EndDate datetime)
returns integer
-- 返回 [@StartDate , @EndDate] 之间周一 到 周日的个数 周日是当周的最后一天
begin
-- @Weekday: 1: Monday , ... ,7: Sunday
return datediff(week,@StartDate,@EndDate)
+ case when (@@Datefirst + datepart(weekday,@StartDate)) % 7
+ case when (@@Datefirst + datepart(weekday,@StartDate)) % 7 = 0
then 7
else
0
end > @Weekday % 7 + 1
then 0
else 1
end
- case when (@@Datefirst + datepart(weekday,@EndDate)) % 7
+ case when (@@Datefirst + datepart(weekday,@EndDate)) % 7 = 0
then 7
else 0
end >= @Weekday % 7 + 1
then
0
else
1
end
/* test:
declare @b datetime
declare @e datetime
set @b = '2004-01-29'
set @e = '2004-09-05'
select @b as BeginDate ,@e as EndDate
,dbo.udf_WeekdayDiff(1,@b,@e) as CountOfMonday
,dbo.udf_WeekdayDiff(2,@b,@e) as CountOfTuesday
,dbo.udf_WeekdayDiff(3,@b,@e) as CountOfWednesday
,dbo.udf_WeekdayDiff(4,@b,@e) as CountOfThursday
,dbo.udf_WeekdayDiff(5,@b,@e) as CountOfFriday
,dbo.udf_WeekdayDiff(6,@b,@e) as CountOfSaturday
,dbo.udf_WeekdayDiff(7,@b,@e) as CountOfSunday
*/
end
go
create function udf_WeekdayID(@Date datetime)
returns integer
-- 返回 @Date 是 Monday 返回 1, ... ,是 Sunday 返回 1
begin
--1: Monday , ... ,7: Sunday
return (@@Datefirst + datepart(weekday,@Date)) % 7
+ case when (@@Datefirst + datepart(weekday,@Date)) % 7 < 2
then 6
else
-1
end
end
go
create function udf_DayOfQuarter(@Date datetime)
-- 返回 @Date 是所在季度的第几天
returns integer
as
begin
/*
declare @date datetime
set @date = '2004-4-1'
--*/
return datediff(day
,dateadd(Quarter,datediff(Quarter,0,@Date),0)
,@Date
) + 1
end
go
create function udf_DaysOfQuarterByDate(@Date datetime)
-- 返回 @Date 所在季度的天数
returns integer
begin
/*
declare @date datetime
set @date = '2004-4-1'
--*/
return datediff(day
,dateadd(Quarter,datediff(Quarter,0,@Date),0)
,dateadd(Quarter,datediff(Quarter,0,@Date) + 1,0)
)
end
go
create function udf_NextWorkDate(@Date datetime)
returns datetime
-- 返回 @Date 的下一个工作日
begin
/*
declare @i int
set @i = 3
declare @Date datetime
set @Date = '2005-01-02'
-- */
return case when (@@Datefirst + datepart(weekday,@Date)) % 7 = 6 -- Friday
then dateadd(day,3,@Date)
when (@@Datefirst + datepart(weekday,@Date)) % 7 = 0 -- saturday
then dateadd(day,2,@Date)
else
dateadd(day,1,@Date)
end
end
go
create function udf_PreviousWorkDate(@Date datetime)
returns datetime
-- 返回 @Date 的上一个工作日
begin
/*
declare @i int
set @i = 3
declare @Date datetime
set @Date = '2005-01-02'
-- */
return case when (@@Datefirst + datepart(weekday,@Date)) % 7 = 2 -- Monday
then dateadd(day,-3,@Date)
when (@@Datefirst + datepart(weekday,@Date)) % 7 = 1 -- Sunday
then dateadd(day,-2,@Date)
else
dateadd(day,-1,@Date)
end
end
go
create function udf_WorkDateAdd(@i integer,@Date datetime)
returns datetime
-- 返回 @Date 加上一段 @i 个工作日的新值
begin
declare @ int
set @ = 0
while @ < abs(@i)
begin
set @Date = case when @i >= 0
then --dbo.udf_nextworkdate(@Date)
case when (@@Datefirst + datepart(weekday,@Date)) % 7 = 6 -- Friday
then @date + 3 --dateadd(day,3,@Date)
when (@@Datefirst + datepart(weekday,@Date)) % 7 = 0 -- saturday
then @date + 2 --dateadd(day,2,@Date)
else
@date + 1 --dateadd(day,1,@Date)
end
else
--dbo.udf_previousworkdate(@Date)
case when (@@Datefirst + datepart(weekday,@Date)) % 7 = 2 -- Monday
then @date - 3 --dateadd(day,-3,@Date)
when (@@Datefirst + datepart(weekday,@Date)) % 7 = 1 -- Sunday
then @date - 2 --dateadd(day,-2,@Date)
else
@date - 1 --dateadd(day,-1,@Date)
end
end
set @ = @ + 1
end
return @Date
end
go
create function udf_GetStar (@ datetime)
returns varchar(100)
-- 返回日期所属星座,如果有静态的 星座对照码表 直接在查询中 join 效率相对更高
begin
return
(
--declare @ datetime
--set @ = getdate()
select max(star)
from
(
select '魔羯座' as star,1 as [month],1 as [day]
union all select '水瓶座',1,20
union all select '双鱼座',2,19
union all select '牡羊座',3,21
union all select '金牛座',4,20
union all select '双子座',5,21
union all select '巨蟹座',6,22
union all select '狮子座',7,23
union all select '处女座',8,23
union all select '天秤座',9,23
union all select '天蝎座',10,24
union all select '射手座',11,22
union all select '魔羯座',12,22
) stars
where dateadd(month,[month] - 1,dateadd(year,year(@) - year(0),0)) + [day] - 1 =
(
select max(dateadd(month,[month] - 1,dateadd(year,year(@) - year(0),0)) + [day] - 1)
from (
select '魔羯座' as star,1 as [month],1 as [day]
union all select '水瓶座',1,20
union all select '双鱼座',2,19
union all select '牡羊座',3,21
union all select '金牛座',4,20
union all select '双子座',5,21
union all select '巨蟹座',6,22
union all select '狮子座',7,23
union all select '处女座',8,23
union all select '天秤座',9,23
union all select '天蝎座',10,24
union all select '射手座',11,22
union all select '魔羯座',12,22
) stars
where @ >= dateadd(month,[month] - 1,dateadd(year,year(@) - year(0),0)) + [day] - 1
)
)
end
go
-- 注意这里用 40 足够了,因为每个月至多才 31 天
select a.birthdate,b.star
from employees a
left join
(
select a.*,isnull(b.month,12) as m,isnull(b.day,31) as d
from stars a
left join stars b
on a.month * 40 + a.day < b.month * 40 + b.day
and b.month * 40 + b.day = (select min(month * 40 + day) from stars where month * 40 + day > a.month * 40 + a.day)
) b
on month(a.birthdate) * 40 + day(a.birthdate) >= b.month * 40 + b.day
and month(a.birthdate) * 40 + day(a.birthdate) < b.m * 40 + b.d
select e.birthdate,a.star
from employees e
left join stars a
on month(e.birthdate) * 40 + day(e.birthdate) >= a.month * 40 + a.day
left join stars b
on a.month * 40 + a.day < b.month * 40 + b.day
and b.month * 40 + b.day = (select min(month * 40 + day) from stars where month * 40 + day > a.month * 40 + a.day)
where month(e.birthdate) * 40 + day(e.birthdate) < isnull(b.month * 40 + b.day,999)
select *
from stars a
left join stars b
on a.month * 40 + a.day < b.month * 40 + b.day
and b.month * 40 + b.day = (select min(month * 40 + day) from stars where month * 40 + day > a.month * 40 + a.day)
go
--周历
declare @ datetime
set @ = getdate()
select @ +
case when (@@Datefirst + datepart(weekday,@)) % 7 = 0 --周六
then -5
when (@@Datefirst + datepart(weekday,@)) % 7 = 1 --周日(七)
then -6
when (@@Datefirst + datepart(weekday,@)) % 7 between 2 and 6 --周一至周五
then 2 - (@@Datefirst + datepart(weekday,@)) % 7
end + N.i
from
(
select 0 as i
union all select 1 union all select 2 union all select 3
union all select 4 union all select 5 union all select 6
union all select 7
) N
//csc noname1.cs
// 装配脑袋 作 C# 计算任意时段内的 之间周一 到 周日的个数 周日是当周的最后一天
using System;
public class Class1
{
static void Main(string[] args)
{
System.Console.WriteLine("Hello World");
System.DateTime Start = System.DateTime.Parse("2005-3-6");
System.DateTime End = System.DateTime.Parse("2005-3-7");
System.Console.WriteLine(WeekdaysIn(DayOfWeek.Monday,Start,End)); //周一
System.Console.WriteLine(WeekdaysIn(DayOfWeek.Tuesday,Start,End));
System.Console.WriteLine(WeekdaysIn(DayOfWeek.Wednesday,Start,End));
System.Console.WriteLine(WeekdaysIn(DayOfWeek.Thursday,Start,End));
System.Console.WriteLine(WeekdaysIn(DayOfWeek.Friday,Start,End));
System.Console.WriteLine(WeekdaysIn(DayOfWeek.Saturday,Start,End));
System.Console.WriteLine(WeekdaysIn(DayOfWeek.Sunday,Start,End)); //周七(日)
System.Console.ReadLine();
}
private static int WeekdaysIn(DayOfWeek Weekday, DateTime StartDate, DateTime EndDate)
{
int d = (EndDate - StartDate).Days;
int w = d / 7;
int offset = (int) StartDate.DayOfWeek + d % 7;
if ((int) Weekday < (int) StartDate.DayOfWeek)
{
Weekday = (DayOfWeek)((int) Weekday + 7);
}
if ((int)Weekday <= offset)
{
w ++;
}
return w;
}
}
