将日期格式更改为较短的日期

Question

我们有以下脚本：

WITH 
SET [Last56Days] as
 TAIL ( [Date].[Date - Calendar Month].[Calendar Day].members, 56 )
MEMBER [Measures].[DateValue] as
 [Date].[Date - Calendar Month].CURRENTMEMBER.member_value, format_string = "short date"
MEMBER [Measures].[DateValue2] as
 [Date].[Date - Calendar Month].CURRENTMEMBER.member_value, format_string = "dd/mm/yyyy"
SELECT
    { [Measures].[DateValue], [Measures].[DateValue2]} ON COLUMNS,
    Hierarchize ({ [Last56Days] } ) ON ROWS
FROM [Our Cube]

它返回这个：

我能否以某种方式更改日期格式，使日期类似于“2014 年 2 月 9 日”，即更短？

Answer 1

从 cmets 中，我认为问题在于您有整数日期键和一个名称列，该列在 Analysis Services 中始终是字符串类型。这些不适用于需要日期的日期格式（或包含自 1900 年 1 月 1 日以来的天数的双精度数，以及一天中时间的分数，即上午 8:30 将是 8.5/24）。

因此，您可以使用

MEMBER [Measures].[Date as int] as
       [Date].[Date - Calendar Month].CURRENTMEMBER.Properties('Key0', Typed)
MEMBER [Measures].[Date Year] as
       Fix([Measures].[Date as int] / 10000)
MEMBER [Measures].[Date Month] as
       Fix(([Measures].[Date as int] - [Measures].[Date Year] * 10000) / 100)
MEMBER [Measures].[Date Day] as
       [Measures].[Date as int] - [Measures].[Date Year] * 10000 - [Measures].[Date Month] * 100
MEMBER [Measures].[DateValue] as
       // convert it to Date data type and use a format string on that:
       DateSerial([Measures].[Date Year], [Measures].[Date Month], [Measures].[Date Day]),
       format_string = 'dd/mm/yyyy'

参见the documentation of Properties 和list of VBA functions like Fix and DateSerial。

看我上面代码的复杂度，使用字符串逻辑而不是整数逻辑可能会更好，即。 e.将typed 参数省略到Properties，然后使用Left、Mid 和Right 计算年、月和日，并将其转换为整数。但是“其中的细节留给读者作为练习”。

Answer 2

尝试以下格式之一：

 =Format(Fields!myDateTime.Value, "M/d/yy") ... 6/15/09 
 =Format(Fields!myDateTime.Value, "M/d/yyyy h:mmtt") ... 6/15/2009 2:45PM 
 =Format(Fields!myDateTime.Value, "MM/dd/yy HH:mm") ... 06/15/09 14:45 
 =Format(Fields!myDateTime.Value, "MMM d, yyyy") ... Jun 15, 2009 
 =Format(Fields!myDateTime.Value, "Short Date") ... 6/15/2009 
 =Format(Fields!myDateTime.Value, "Long Date") ... Monday, June 15, 2009 
 =Format(Fields!myDateTime.Value, "ddd dd/MM/yyyy") ... Mon 04/04/2011 

Source Link。希望对你有帮助

编辑 对不起！最后没看到评论。无论如何，我猜你已经看过这个Link，但你也许可以尝试像这样使用 format_string：

format_string = "dd/mmm/yyyy"