本文介绍了序列化的DateTime,恕不毫秒,格林尼治标准​​时间的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我已经使用XSD文件作为输入创建一个C#类文件。我的一个特性是这样的:

I have created a C# class file by using a XSD-file as an input. One of my properties look like this:

 private System.DateTime timeField;

 [System.Xml.Serialization.XmlElementAttribute(DataType="time")]
 public System.DateTime Time {
     get {
         return this.timeField;
     }
     set {
         this.timeField = value;
     }
 }

在序列化,该文件的内容现在看起来是这样的:

When serialized, the contents of the file now looks like this:

<Time>14:04:02.1661975+02:00</Time>

是否有可能,有对物业XMLATTRIBUTES,有它使没有毫秒,格林尼治标准​​时间值也是这样吗?

Is it possible, with XmlAttributes on the property, to have it render without the milliseconds and the GMT-value like this?

<Time>14:04:02</Time>

这是可能的,或者我需要破解在一起某种XSL / XPath的替换魔术类已系列化后?

Is this possible, or do i need to hack together some sort of xsl/xpath-replace-magic after the class has been serialized?

有不是解决改变对象到字符串,因为它用于像在应用程序的其余部分DateTime和允许我们通过使用XmlSerializer.Serialize创建从对象一个xml-重新presentation ()方法。

It is not a solution to changing the object to String, because it is used like a DateTime in the rest of the application and allows us to create an xml-representation from an object by using the XmlSerializer.Serialize() method.

我需要从外地删除多余信息的原因是接收系统不符合W3C标准的标准时间数据类型。

The reason I need to remove the extra info from the field is that the receiving system does not conform to the w3c-standards for the time datatype.

推荐答案

您可以创建一个字符串属性,做翻译到/从你的timeField领域,并把序列化的属性上而不是真实的日期时间属性,其余应用程序使用。

You could create a string property that does the translation to/from your timeField field and put the serialization attribute on that instead the the real DateTime property that the rest of the application uses.

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09-02 11:30