我想根据例如22/01/2015或2016-12-15的模式格式化日期。
在.NET Framework中,我们有DateTime - > ToString()方法接受格式作为参数甚至是string.Format,它也是如此。
MS SQL Server 2008上是否有基于参数模式格式化日期的函数?
答案 0 :(得分:7)
我在SQL 2012中遇到了类似的问题,他们有一个很棒的函数叫做格式,你可以传递一个日期和你需要的格式
SELECT FORMAT(GETDATE(), 'dd/mm/yyy');
在SQL Server 2008中没有类似的内容。我创建了一个简单的函数来完成类似的工作,它可能需要调整但不是一个坏基础。
-- =============================================
-- Author: Luke Mc Redmond
-- =============================================
CREATE FUNCTION [dbo].[FormatDate]
(
-- Add the parameters for the function here
@Date DATETIME,
@Format NVARCHAR(50)
)
RETURNS NVARCHAR(50)
AS
BEGIN
DECLARE @ResultVar nvarchar(50) = UPPER(@Format);
DECLARE @d nvarchar(20) = CONVERT(nvarchar(20), DATEPART(day, @Date));
DECLARE @dd nvarchar(20) = CONVERT(nvarchar(20), DATEPART(day, @Date));
DECLARE @day nvarchar(20) = DATENAME(weekday, @Date);
DECLARE @ddd nvarchar(20) = DATENAME(weekday, @Date)+' '+CONVERT(nvarchar(5), DATEPART(day, @Date));
DECLARE @m nvarchar(20) = CONVERT(nvarchar, DATEPART(month, @Date));
DECLARE @mm nvarchar(20) = CONVERT(nvarchar, DATEPART(month, @Date));
DECLARE @mmm nvarchar(20) = CONVERT(VARCHAR(3), DATENAME(month, @Date), 100);
DECLARE @month nvarchar(20) = DATENAME(month, @Date);
DECLARE @y nvarchar(20) = CONVERT(nvarchar, DATEPART(year, @Date));
DECLARE @yy nvarchar(20) = RIGHT(CONVERT(nvarchar, DATEPART(year, GETDATE())),2);
DECLARE @yyy nvarchar(20) = CONVERT(nvarchar, DATEPART(year, @Date));
DECLARE @yyyy nvarchar(20) = CONVERT(nvarchar, DATEPART(year, @Date));
DECLARE @year nvarchar(20) = CONVERT(nvarchar, DATEPART(year, @Date));
SELECT @ResultVar = CASE WHEN CHARINDEX('DAY',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS ,'DAY' collate Latin1_General_CS_AS ,@day)
WHEN CHARINDEX('DDD',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS ,'DDD' collate Latin1_General_CS_AS ,@ddd)
WHEN CHARINDEX('DD',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS ,'DD' collate Latin1_General_CS_AS ,@dd)
WHEN CHARINDEX('D',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS,'D' collate Latin1_General_CS_AS ,@d) END;
SELECT @ResultVar = REPLACE(@ResultVar collate Latin1_General_CS_AS,'Monday' collate Latin1_General_CS_AS ,'monday')
SELECT @ResultVar = CASE WHEN CHARINDEX('MONTH',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS,'MONTH' collate Latin1_General_CS_AS,@month)
WHEN CHARINDEX('MMM',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS ,'MMM' collate Latin1_General_CS_AS,@mmm)
WHEN CHARINDEX('MM',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS,'MM' collate Latin1_General_CS_AS,@mm)
WHEN CHARINDEX('M',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS,'M' collate Latin1_General_CS_AS,@m) END;
SELECT @ResultVar = CASE WHEN CHARINDEX('YEAR',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS,'YEAR' collate Latin1_General_CS_AS ,@year)
WHEN CHARINDEX('YYYY',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS,'YYYY' collate Latin1_General_CS_AS ,@yyyy)
WHEN CHARINDEX('YYY',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS ,'YYY' collate Latin1_General_CS_AS ,@yyy)
WHEN CHARINDEX('YY',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS ,'YY' collate Latin1_General_CS_AS ,@yy)
WHEN CHARINDEX('Y',@ResultVar) > 0 THEN REPLACE(@ResultVar collate Latin1_General_CS_AS,'Y' collate Latin1_General_CS_AS ,@y) END;
SELECT @ResultVar = REPLACE(@ResultVar collate Latin1_General_CS_AS,'monday' collate Latin1_General_CS_AS ,'Monday')
RETURN @ResultVar
END;
GO
希望这有帮助!
答案 1 :(得分:0)
将CONVERT函数与style参数一起使用。它全部列在这里:https://msdn.microsoft.com/en-us/library/ms187928.aspx