如何在 c# 中序列化一个序列(xs 序列,maxOccurs 不同于 0)

How to serialize a sequence in c# (xs sequence with maxOccurs different that 0)

我在使用 C# 序列化数组以符合 XSD 时遇到问题。

我需要序列化一个数组,其中每个 child 的每个 属性 都将列在一个列表中,以符合定义如下的 XSD:

<xs:sequence minOccurs="0" maxOccurs="unbounded">
    <xs:element name="Id" type="xs:string"/>
    <xs:element name="Value" type="xs:string"/>
</xs:sequence>

例如,我有这个:

<ServerRequest xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <Time>11-08-15 08:27:31</Time>
  <RequestContent>
    <Id>myId1</Id>
    <Value>myValue1</Value>
  </RequestContent>
  <RequestContent>
    <Id>myId2</Id>
    <Value>myValue2</Value>
  </RequestContent>
</ServerRequest>

而我需要的是:

<ServerRequest xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <Time>11-08-15 08:27:31</Time>
  <Id>myId1</Id>
  <Value>myValue1</Value>
  <Id>myId2</Id>
  <Value>myValue2</Value>
</ServerRequest>

这是我的代码示例:

class Program
{
    static void Main(string[] args)
    {
        ServerRequest test = new ServerRequest();
        test.Time = DateTime.UtcNow.ToString();
        test.RequestContent = new List<ServerRequestContent>()
        {
            new ServerRequestContent() { Id = "myId1", Value = "myValue1"},
            new ServerRequestContent() { Id = "myId2", Value = "myValue2"}
        };

        using (StringWriter sw = new StringWriter())
        {
            XmlSerializer xmlSerializer = new XmlSerializer(typeof(ServerRequest));
            xmlSerializer.Serialize(sw, test);

            Console.WriteLine(sw.ToString());
        }
    }
}


[XmlRoot("ServerRequest")]
public class ServerRequest
{
    [XmlElement()]
    public string Time { get; set; }

    [XmlElement]
    public List<ServerRequestContent> RequestContent { get; set; }
}

public class ServerRequestContent
{
    [XmlElement()]
    public string Id { get; set; }

    [XmlElement()]
    public string Value { get; set; }
}

我已经尝试了几个小时,但仍然找不到解决方案。到目前为止我发现的最好的东西是:Serialize Array without root element,但我必须在从 XSD 生成的 C# classes 中更改很多东西,我真的不想要到.

感谢您的帮助

解法:

实施 IXmlSerializable 可能是处理此问题的最简单方法。 ServerRequest class 现在看起来像这样:

[XmlRoot("ServerRequest")]
public class ServerRequest : IXmlSerializable
{
    [XmlElement()]
    public string Time { get; set; }

    [XmlElement]
    public List<ServerRequestContent> RequestContent { get; set; }

    #region IXmlSerializable

    public XmlSchema GetSchema()
    {
        return null;
    }

    public void ReadXml(XmlReader reader)
    {
        //...
    }

    public void WriteXml(XmlWriter writer)
    {
        writer.WriteElementString("Time", Time);

        foreach (ServerRequestContent content in RequestContent)
        {
            writer.WriteElementString("Id", content.Id);
            writer.WriteElementString("Value", content.Value);
        }
    }

    #endregion
}

您应该能够通过在 ServerRequest 上实施 IXmlSerializable 并覆盖 WriteXml(XmlWriter) 来实现此目的。另见

或者,虽然可以说是 hack,但您可以通过 XSLT 转换传递最终输出。例如:

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:template match="/ServerRequest">
        <ServerRequest xmlns:xsd="http://www.w3.org/2001/XMLSchema" 
                       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
            <Time><xsl:value-of select="Time"/></Time>
            <xsl:for-each select="RequestContent">
                <Id><xsl:value-of select="Id"/></Id>
                <Value><xsl:value-of select="Value"/></Value>
            </xsl:for-each>
        </ServerRequest>
    </xsl:template>
</xsl:stylesheet>