将XML字符串反序列化为复杂类型 [英] Deserialize XML string to complex type

问题描述

我的Xml（我无法更改）：

My Xml (which I can't change):

<result>
    <type>MAZDA</type>
    <make>RX-8</type>
    <country>JAPAN</country>
</result>

我的模型：

[Serializable, XmlRoot("result")]
public class VehicleDetails
{
    public string Type { get; set; }
    public string Make { get; set; }
    public string Country { get; set; }
}

反序列化此XML的工作符合预期，但我想更改 Country 属性转换为复杂类型，例如：

de-serializing this XML works as expected but I want to change the Country property to a complex type, like so:

public Country Country { get; set; }

并在 Country中输入国家名称 JAPAN。名称属性，有什么想法吗？

and put the country name, "JAPAN", in the Country.Name property, any ideas?

推荐答案

您可以装饰 Name <您的 Country 类的/ code>属性和 [XmlText] 属性，例如：

You could decorate the Name property of your Country class with the [XmlText] attribute like this:

[XmlRoot("result")]
public class VehicleDetails
{
    public string Type { get; set; }
    public string Make { get; set; }
    public Country Country { get; set; }
}

public class Country
{
    [XmlText]
    public string Name { get; set; }
}

还要注意，我已经摆脱了 [可序列化] 属性。它对XML序列化没有用。此属性用于二进制/远程序列化。

Also notice that I have gotten rid of the [Serializable] attribute. It is useless for XML serialization. This attribute is used for binary/remoting serialization.

这是一个完整的示例，将按预期打印 JAPAN ：

And here's a full example that will print JAPAN as expected:

using System;
using System.IO;
using System.Xml;
using System.Xml.Serialization;

[XmlRoot("result")]
public class VehicleDetails
{
    public string Type { get; set; }
    public string Make { get; set; }
    public Country Country { get; set; }
}

public class Country
{
    [XmlText]
    public string Name { get; set; }
}

class Program
{
    static void Main()
    {
        var serializer = new XmlSerializer(typeof(VehicleDetails));
        var xml = 
        @"<result>
            <Type>MAZDA</Type>
            <Make>RX-8</Make>
            <Country>JAPAN</Country>
        </result>";
        using (var reader = new StringReader(xml))
        using (var xmlReader = XmlReader.Create(reader))
        {
            var result = (VehicleDetails)serializer.Deserialize(xmlReader);
            Console.WriteLine(result.Country.Name);
        }
    }
}

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