如何使用DATEDIFF返回SQL Server 2005中年，月，日两个日期之间的差异
DATEDIFF（日期，日期）
如何结果：2年3个月10天
任何人都可以完成这个t-sql吗？
ALTER FUNCTION [dbo].[gatYMD](@dstart VARCHAR(50), @dend VARCHAR(50))
RETURNS VARCHAR(50) AS
BEGIN
DECLARE @yy INT
DECLARE @mm INT
DECLARE @getmm INT
DECLARE @dd INT
SET @yy = DATEDIFF(yy, @dstart, @dend)
SET @mm = DATEDIFF(mm, @dstart, @dend)
SET @dd = DATEDIFF(dd, @dstart, @dend)
SET @getmm = ABS(DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend))
RETURN (
Convert(varchar(10),@yy) + 'year' + Convert(varchar(10),@mm) + 'month' + Convert(varchar(10),@dd) + 'day'
)
END
答案 0 :(得分：6)
这是我对Eric功能的解决方案：
DECLARE @getmm INT
DECLARE @getdd INT
SET @yy = DATEDIFF(yy, @dstart, @dend)
SET @mm = DATEDIFF(mm, @dstart, @dend)
SET @dd = DATEDIFF(dd, @dstart, @dend)
SET @getmm = ABS(DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend))
SET @getdd = ABS(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend), DATEADD(yy, @yy, @dstart)), @dend))
RETURN (
Convert(varchar(10),@yy) + 'year' + Convert(varchar(10),@getmm) + 'month' + Convert(varchar(10),@getdd) + 'day'
)
如果开始日期在结束日期之后，请使用ABS来处理。
此：
WITH ex_table AS (
SELECT '2007-01-01' 'birthdatetime',
'2009-03-29' 'visitdatetime')
SELECT CAST(DATEDIFF(yy, t.birthdatetime, t.visitdatetime) AS varchar(4)) +' year '+
CAST(DATEDIFF(mm, DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime), t.visitdatetime) AS varchar(2)) +' month '+
CAST(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime), t.visitdatetime), DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime)), t.visitdatetime) AS varchar(2)) +' day' AS result
FROM ex_table t
..或用于SQL Server 2000及之前的非CTE：
SELECT CAST(DATEDIFF(yy, t.birthdatetime, t.visitdatetime) AS varchar(4)) +' year '+
CAST(DATEDIFF(mm, DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime), t.visitdatetime) AS varchar(2)) +' month '+
CAST(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime), t.visitdatetime), DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime)), t.visitdatetime) AS varchar(2)) +' day' AS result
FROM (SELECT '2007-01-01' 'birthdatetime',
'2009-03-29' 'visitdatetime') t
...将返回：
result
----------------------
2 year 2 month 28 day
参考：DATEDIFF
答案 1 :(得分：3)
它适用于某些情况，但是当您从2011-01-13（dateterminated）减去2007-10-15（DateHired）等日期时，它会给你一个负数，当abs不在它周围但放绝对不会解决它，因为那时年月数不正确。
答案 2 :(得分：3)
ALTER FUNCTION [dbo].[gatYMD](@dstart VARCHAR(50), @dend VARCHAR(50))
RETURNS VARCHAR(50) AS
BEGIN
DECLARE @yy INT
DECLARE @mm INT
DECLARE @getdd INT
DECLARE @dd INT
SET @yy = DATEDIFF(yy, @dstart, @dend)
SET @mm = DATEDIFF(mm, @dstart, @dend) - (12 * @yy)
SET @dd = ABS(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend), DATEADD(yy, @yy, @dstart)), @dend))
Return (Convert(varchar(10),@yy) + ' year ' + Convert(varchar(10),@mm) + ' month ' + Convert(varchar(10),@dd) + ' day ')
END
答案 3 :(得分：2)
检查此页面... http://www.sqlteam.com/article/datediff-function-demystified
创建此功能：
CREATE FUNCTION dbo.fnYearsApart
(
@FromDate DATETIME,
@ToDate DATETIME
)
RETURNS INT
AS
BEGIN
RETURN CASE
WHEN @FromDate > @ToDate THEN NULL
WHEN DATEPART(day, @FromDate) > DATEPART(day, @ToDate) THEN DATEDIFF(month, @FromDate, @ToDate) - 1
ELSE DATEDIFF(month, @FromDate, @ToDate)
END / 12
END
和
CREATE FUNCTION dbo.fnMonthsApart
(
@FromDate DATETIME,
@ToDate DATETIME
)
RETURNS INT
AS
BEGIN
RETURN CASE
WHEN @FromDate > @ToDate THEN NULL
WHEN DATEPART(day, @FromDate) > DATEPART(day, @ToDate) THEN DATEDIFF(month, @FromDate, @ToDate) - 1
ELSE DATEDIFF(month, @FromDate, @ToDate)
END
END
最后
ALTER FUNCTION [dbo].[gatYMD](@dstart VARCHAR(50), @dend VARCHAR(50))
RETURNS VARCHAR(50) AS
BEGIN
DECLARE @yy INT
DECLARE @mm INT
DECLARE @dd INT
DECLARE @getmm INT
DECLARE @getdd INT
SET @yy = dbo.fnYearsApart(@dstart, @dend) --DATEDIFF(yy, @dstart, @dend)
SET @mm = dbo.fnMonthsApart(@dstart, @dend) --DATEDIFF(mm, @dstart, @dend)
SET @dd = DATEDIFF(dd, @dstart, @dend)
SET @getmm = ABS(DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend))
SET @getdd = ABS(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend), DATEADD(yy, @yy, @dstart)), @dend))
RETURN (
Convert(varchar(10),@yy) + ' años, ' + Convert(varchar(10),@getmm) + ' meses, ' + Convert(varchar(10),@getdd) + ' días'
)
END
巨星！
答案 4 :(得分：2)
创建此功能，它会提供完全日期差异，如年月日
Create function get_Exact_Date_diff(@date smalldatetime,@date2 smalldatetime)
returns varchar(50)
as
begin
declare @date3 smalldatetime
Declare @month int,@year int,@day int
if @date>@date2
begin
set @date3=@date2
set @date2=@date
set @date=@date3
end
SELECT @month=datediff (MONTH,@date,@date2)
if dateadd(month,@month,@date) >@date2
begin
set @month=@month-1
end
set @day=DATEDIFF(day,dateadd(month,@month,@date),@date2)
set @year=@month/12
set @month=@month % 12
return (case when @year=0 then '' when @year=1 then convert(varchar(50),@year ) + ' year ' when @year>1 then convert(varchar(50),@year ) + ' years ' end)
+ (case when @month=0 then '' when @month=1 then convert(varchar(50),@month ) + ' month ' when @month>1 then convert(varchar(50),@month ) + ' months ' end)
+ (case when @day=0 then '' when @day=1 then convert(varchar(50),@day ) + ' day ' when @day>1 then convert(varchar(50),@day ) + ' days ' end)
end
答案 5 :(得分：2)
我知道这里已经有一些答案，但我想我会添加我想出的内容（至少对我而言），这似乎很容易理解：
CREATE FUNCTION dbo.fn_DateDiff_YMDMHS
(
@Startdate as datetime2(0),
@Enddate as datetime2(0)
)
RETURNS TABLE
AS
RETURN
(
select
TotalYears [Years],
datediff(month, dateadd(Year, TotalYears, @Startdate), @Enddate) Months,
datediff(day, dateadd(month, TotalMonths, @Startdate), @Enddate) [Days],
datediff(hour, dateadd(day, TotalDays, @Startdate), @Enddate) [Hours],
datediff(minute, dateadd(hour, TotalHours, @Startdate), @Enddate) [Minutes],
datediff(second, dateadd(minute, TotalMinutes, @Startdate), @Enddate) [Seconds]
from (
select
datediff(SECOND, @Startdate, @Enddate) TotalSeconds,
datediff(minute, @Startdate, @Enddate) TotalMinutes,
datediff(hour, @Startdate, @Enddate) TotalHours,
datediff(day, @Startdate, @Enddate) TotalDays,
datediff(month, @Startdate, @Enddate) TotalMonths,
datediff(year, @Startdate, @Enddate) TotalYears) DateDiffs
)
然后当你打电话：
select * from dbo.fn_DateDiff_YMDMHS('1900-01-01 00:00:00', '1910-10-05 03:01:02')
你会得到这个：
Years Months Days Hours Minutes Seconds
10 9 4 3 1 2
显然你可以将它更改为格式化输出并改为使用标量变量，但我会留给你:-)
编辑：
我最后还需要做一段时间的功能才能返回“5年零2天前”这样的格式
CREATE FUNCTION fn_DateDiff_YMDMHS_String
(
@StartDate datetime2(0),
@EndDate datetime2(0),
@OutputYears bit = 1,
@OutputMonths bit = 1,
@OutputDays bit = 1,
@OutputHours bit = 0,
@OutputMinutes bit = 0,
@OutputSeconds bit = 0,
@OutputSuffix bit = 0
)
RETURNS varchar(256)
AS
BEGIN
DECLARE @Output varchar(256) = ''
declare @Years int, @Months int, @Days int, @Hours int, @Minutes int, @Seconds int
select
@Years = case when @OutputYears = 1 then Years else 0 end,
@Months = case when @OutputMonths = 1 then Months else 0 end,
@Days = case when @OutputDays = 1 then Days else 0 end,
@Hours = case when @OutputHours = 1 then Hours else 0 end,
@Minutes = case when @OutputMinutes = 1 then Minutes else 0 end,
@Seconds = case when @OutputSeconds = 1 then Seconds else 0 end
from dbo.fn_DateDiff_YMDMHS(@StartDate, @EndDate)
declare @and varchar(5) = ''
if @OutputYears = 1 and @Years > 0
begin
set @Output = @Output + cast(@Years as varchar(4)) + ' year'
if @Years > 1 set @Output = @Output + 's ' else set @Output = @Output + ' '
if @Months > 0 and @Days + @Hours + @Minutes + @Seconds = 0 set @and = 'and '
end
if @OutputMonths = 1 and @Months > 0
begin
set @Output = @Output + @and + cast(@Months as varchar(2)) + ' month'
if @Months > 1 set @Output = @Output + 's ' else set @Output = @Output + ' '
if @Days > 0 and @Hours + @Minutes + @Seconds = 0 set @and = 'and '
end
if @OutputDays = 1 and @Days > 0
begin
set @Output = @Output + @and + cast(@Days as varchar(2)) + ' day'
if @Days > 1 set @Output = @Output + 's ' else set @Output = @Output + ' '
if @Hours > 0 and @Minutes + @Seconds = 0 set @and = 'and '
end
if @OutputHours = 1 and @Hours > 0
begin
set @Output = @Output + @and + cast(@Hours as varchar(2)) + ' hour'
if @Hours > 1 set @Output = @Output + 's ' else set @Output = @Output + ' '
if @Minutes > 0 and @Seconds = 0 set @and = 'and '
end
if @OutputMinutes = 1 and @Minutes > 0
begin
set @Output = @Output + @and + cast(@Minutes as varchar(2)) + ' minute'
if @Minutes > 1 set @Output = @Output + 's ' else set @Output = @Output + ' '
if @Seconds > 0 set @and = 'and '
end
if @OutputSeconds = 1 and @Seconds > 0
begin
set @Output = @Output + @and + cast(@Seconds as varchar(2)) + ' second'
if @Seconds > 1 set @Output = @Output + 's ' else set @Output = @Output + ' '
end
if @OutputSuffix = 1
begin
if @StartDate < @EndDate
begin
set @Output = @Output + 'ago'
end
else
begin
set @Output = 'in ' + @Output
end
end
RETURN @Output
END
以下是2个例子
select dbo.fn_DateDiff_YMDMHS_String('2000-01-01 00:00:00', '2007-09-19 14:39:53', 1, 1, 1, 1,1,1, 1)
7 years 8 months 18 days 8 hours 39 minutes and 53 seconds ago
和
select dbo.fn_DateDiff_YMDMHS_String('2000-01-01 00:00:00', '2007-09-19 14:39:53', 1, 1, 1, 0,0,0, 1)
7 years 8 months and 18 days ago
我希望这对未来的某个人有用，因为我在搜索时找不到这样的东西（虽然可能只是那些日子之一）我愿意改进，因为我知道我是并不总是最紧凑或最快的代码编程器： - ）
此致
利安
答案 6 :(得分：1)
Using ParseName
DECLARE
@ReportBeginDate DATE
SET @ReportBeginDate='2015-01-01';
IF OBJECT_ID('TEMPDB..#tmp_ymd') IS NOT NULL
BEGIN
DROP TABLE #tmp_ymd;
END;
select
cast(cast(datediff(mm,@ReportBeginDate,getdate()) as decimal (10,2))/12 as decimal(10,2)) as YearMonthDec
,cast(datediff(dd,@ReportBeginDate,getdate()) as decimal (10,2)) as DayDec
into #tmp_ymd
select
YearMonthDec
,cast(parsename(YearMonthDec,2) as decimal (10,0)) as yearnum
,cast(cast(parsename(YearMonthDec,1) as decimal (10,0))/100*(12) as numeric) as monthnum
,case when YearMonthDec>=1 then datediff(dd,CONVERT(VARCHAR(25),DATEADD(dd,-(DAY(getdate())-1),getdate()),101),getdate()) else DayDec end as daynum
from #tmp_ymd
答案 7 :(得分：0)
CREATE FUNCTION FindDateDiff(@Date1 date,@Date2 date)
RETURNS TABLE
AS
RETURN
(
SELECT
CALC.Years,CALC.Months,D.Days,
Result = RTRIM(Case When CALC.Years > 0 Then CONCAT(CALC.Years, ' year(s) ') Else '' End
+ Case When CALC.Months > 0 Then CONCAT(CALC.Months, ' month(s) ') Else '' End
+ Case When D.Days > 0 OR (CALC.Years=0 AND CALC.Months=0) Then CONCAT(D.Days, ' day(s)') Else '' End)
FROM (VALUES(IIF(@Date1<@Date2,@Date1,@Date2),IIF(@Date1<@Date2,@Date2,@Date1))) T(StartDate, EndDate)
CROSS APPLY(Select
TempEndYear = Case When ISDATE(CONCAT(YEAR(T.EndDate), FORMAT(T.StartDate,'-MM-dd')))=1 Then CONCAT(YEAR(T.EndDate), FORMAT(T.StartDate,'-MM-dd'))
Else CONCAT(YEAR(T.EndDate),'-02-28') End
) TEY
CROSS APPLY(Select EndYear = Case When TEY.TempEndYear > T.EndDate Then DATEADD(YEAR, -1, TEY.TempEndYear) Else TEY.TempEndYear End) EY
CROSS APPLY(Select
Years = DATEDIFF(YEAR,T.StartDate,EY.EndYear),
Months = DATEDIFF(MONTH,EY.EndYear,T.EndDate)-IIF(DAY(EY.EndYear)>DAY(T.EndDate),1,0)
) CALC
CROSS APPLY(Select Days = DATEDIFF(DAY,DATEADD(MONTH,CALC.Months,DATEADD(YEAR,CALC.Years,T.StartDate)),T.EndDate)) D
)
和
Select * From dbo.FindDateDiff('2015-12-15','2018-12-14')
结果：
答案 8 :(得分：0)
TL; DR无需创建函数即可近似年月。
我在搜索“从datediff获得年月”时发现了这个问题。我正在寻找比上述函数创建解决方案更快（是的，可能更脏）的东西。
我想出了以下行内SQL，它近似年月。对于我自己的目的已经足够了，它可以根据当时用户的粗暴年龄对事件进行计数。
select cast(datediff(event_datetime, dateofbirth)/365.25 as unsigned) as years,
cast((mod(datediff(event_datetime, dateofbirth),365.25))/30.4375 as unsigned) as months
from tablename
;