如何使用动态标记名称将数组序列化为 XML [英] How to serialize an array to XML with dynamic tag names

问题描述

我需要在 C# 上使用动态标记名称将对象数组序列化为 XML.我创建了两个需要序列化为 xml 的类，但我不知道如何创建带有动态名称的标签.

I need to serialize an array of objects to XML on C# with dynamic tag names. I have created two classes that i need to serialize into a xml, but i can't figure out how to create tags with dynamic names.

例如，我有以下课程:

[System.Xml.Serialization.XmlRootAttribute(IsNullable = false)]
public class GeneralInformation
{

    private Info[] addInfoList;

    /// <remarks/>
    [System.Xml.Serialization.XmlArray("InfoList")]
    public Info[] AddInfoList
    {
        get
        {
            return this.addInfoList;
        }
        set
        {
            this.addInfoList = value;
        }
    }
}

public class Info
{
    private string infoMessage;

    /// <remarks/>
    [System.Xml.Serialization.XmlElement("InfoName")]
    public string InfoMessage
    {
        get
        {
            return this.infoMessage;
        }
        set
        {
            this.infoMessage = value;
        }
    }
}

如果我添加一些简单的数据并将其序列化，我会得到:

If I add some simple data and serialize it, I get this:

<?xml version="1.0" encoding="utf-16"?>
<GeneralInformation xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <InfoList>
        <Info>
            <InfoName>Test1</InfoName>
        </Info>
        <Info>
            <InfoName>Test2</InfoName>
        </Info>
        <Info>
            <InfoName>Test3</InfoName>
        </Info>
    </InfoList>
</GeneralInformation>

但是我需要用数组的索引 + 1 来枚举标签Info".这可能吗?结果看起来像这样.

But i need to enumerate the tag "Info" with the index of the array + 1. Is that possible? The result would look like this.

<?xml version="1.0" encoding="utf-16"?>
<GeneralInformation xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <InfoList>
        <Info001>
            <InfoName>Test1</InfoName>
        </Info001>
        <Info002>
            <InfoName>Test2</InfoName>
        </Info002>
        <Info003>
            <InfoName>Test3</InfoName>
        </Info003>
    </InfoList>
</GeneralInformation>

注意我只需要将我的 GeneralInformation 序列化为 XML，而不是反序列化.

Note I only need to serialize my GeneralInformation to XML, not deserialize.

推荐答案

使用 XDocument :

Using XDocument :

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;

namespace ConsoleApplication45
{
    class Program
    {

        static void Main(string[] args)
        {
            string xmlIdent = "<?xml version=\"1.0\" encoding=\"utf-16\"?>" +
                "<GeneralInformation xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\">" +
                "</GeneralInformation>";

            XDocument doc = XDocument.Parse(xmlIdent);

            XElement generalInfo = doc.Root;
            XElement infoList = new XElement("InfoList");
            generalInfo.Add(infoList);

            for (int i = 0; i < 10; i++)
            {
                infoList.Add(new XElement("Infor" + i.ToString("0##"), new XElement("InfoName", "Test" + i.ToString("0##"))));
            }


        }

    }
}

//<?xml version="1.0" encoding="utf-16"?>
//<GeneralInformation xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
//    <InfoList>
//        <Info001>
//            <InfoName>Test1</InfoName>
//        </Info001>
//        <Info002>
//            <InfoName>Test2</InfoName>
//        </Info002>
//        <Info003>
//            <InfoName>Test3</InfoName>
//        </Info003>
//    </InfoList>
//</GeneralInformation>

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