如何使用 C#/LINQ 将 XML 转换为 JSON？

How to convert XML to JSON using C#/LINQ?

提问by Farhad-Taran

I have the following XML file that I need to convert to JSON in the server. Initially I thought I would convert it to a Dictionary and then use the JavaScriptSerializer to turn it into JSON but since each column could have a different value type, I don't think it would work. Has anyone done something similar before in C#/LINQ?

我有以下 XML 文件，需要在服务器中将其转换为 JSON。最初我以为我会将它转换为字典，然后使用 JavaScriptSerializer 将其转换为 JSON，但由于每列可能具有不同的值类型，我认为它行不通。有没有人在 C#/LINQ 中做过类似的事情？

I need to preserve the Value Types(Boolean, String, Integer) of each column.

我需要保留每列的值类型（布尔、字符串、整数）。

I would appreciate any advice on this as Im just starting to work with XML. Thanks.

我将不胜感激，因为我刚刚开始使用 XML。谢谢。

<Columns>
 <Column Name="key1" DataType="Boolean">True</Column>
 <Column Name="key2" DataType="String">Hello World</Column>
 <Column Name="key3" DataType="Integer">999</Column>
</Columns>

采纳答案by Darin Dimitrov

using System;
using System.Linq;
using System.Web.Script.Serialization;
using System.Xml.Linq;

class Program
{
    static void Main()
    {
        var xml = 
        @"<Columns>
          <Column Name=""key1"" DataType=""Boolean"">True</Column>
          <Column Name=""key2"" DataType=""String"">Hello World</Column>
          <Column Name=""key3"" DataType=""Integer"">999</Column>
        </Columns>";
        var dic = XDocument
            .Parse(xml)
            .Descendants("Column")
            .ToDictionary(
                c => c.Attribute("Name").Value, 
                c => c.Value
            );
        var json = new JavaScriptSerializer().Serialize(dic);
        Console.WriteLine(json);
    }
}

produces:

产生：

{"key1":"True","key2":"Hello World","key3":"999"}

Obviously this treats all the values as strings. If you want to keep the underlying type semantics you could do the following:

显然，这将所有值视为字符串。如果您想保留底层类型语义，您可以执行以下操作：

using System;
using System.Linq;
using System.Web.Script.Serialization;
using System.Xml.Linq;

class Program
{
    static void Main()
    {
        var xml = 
        @"<Columns>
          <Column Name=""key1"" DataType=""System.Boolean"">True</Column>
          <Column Name=""key2"" DataType=""System.String"">Hello World</Column>
          <Column Name=""key3"" DataType=""System.Int32"">999</Column>
        </Columns>";
        var dic = XDocument
            .Parse(xml)
            .Descendants("Column")
            .ToDictionary(
                c => c.Attribute("Name").Value, 
                c => Convert.ChangeType(
                    c.Value,
                    typeof(string).Assembly.GetType(c.Attribute("DataType").Value, true)
                )
            );
        var json = new JavaScriptSerializer().Serialize(dic);
        Console.WriteLine(json);
    }
}

produces:

产生：

{"key1":true,"key2":"Hello World","key3":999}

And if you cannot modify the underlying XML structure you will need a custom function that will convert between your custom types and the underlying .NET type:

如果您无法修改底层 XML 结构，您将需要一个自定义函数来在您的自定义类型和底层 .NET 类型之间进行转换：

using System;
using System.Linq;
using System.Web.Script.Serialization;
using System.Xml.Linq;

class Program
{
    static void Main()
    {
        var xml = 
        @"<Columns>
          <Column Name=""key1"" DataType=""Boolean"">True</Column>
          <Column Name=""key2"" DataType=""String"">Hello World</Column>
          <Column Name=""key3"" DataType=""Integer"">999</Column>
        </Columns>";
        var dic = XDocument
            .Parse(xml)
            .Descendants("Column")
            .ToDictionary(
                c => c.Attribute("Name").Value, 
                c => Convert.ChangeType(
                    c.Value, 
                    GetType(c.Attribute("DataType").Value)
                )
            );
        var json = new JavaScriptSerializer().Serialize(dic);
        Console.WriteLine(json);
    }

    private static Type GetType(string type)
    {
        switch (type)
        {
            case "Integer":
                return typeof(int);
            case "String":
                return typeof(string);
            case "Boolean":
                return typeof(bool);
            // TODO: add any other types that you want to support
            default:
                throw new NotSupportedException(
                    string.Format("The type {0} is not supported", type)
                );
        }
    }
}

回答by oopbase

Is it necessary to use LINQ? Otherwise you can try this:

有必要使用LINQ吗？否则你可以试试这个：

XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);
string jsonText = JsonConvert.SerializeXmlNode(doc);

Taken from this post.

取自这篇文章。

回答by Termininja

For deep nesting of XML elements with more and unknown attributes you can use this recursion:

对于具有更多和未知属性的 XML 元素的深层嵌套，您可以使用此递归：

private static string XmlToJson(string xmlString)
{
    return new JavaScriptSerializer().Serialize(GetXmlValues(XElement.Parse(xmlString)));
}

private static Dictionary<string, object> GetXmlValues(XElement xml)
{
    var attr = xml.Attributes().ToDictionary(d => d.Name.LocalName, d => (object)d.Value);
    if (xml.HasElements) attr.Add("_value", xml.Elements().Select(e => GetXmlValues(e)));
    else if (!xml.IsEmpty) attr.Add("_value", xml.Value);

    return new Dictionary<string, object> { { xml.Name.LocalName, attr } };
}

For your example the result will be:

对于您的示例，结果将是：

{
    "Columns":{
        "_value":[
            {
                "Column":{
                    "Name":"key1",
                    "DataType":"Boolean",
                    "_value":"True"
                }
            },
            {
                "Column":{
                    "Name":"key2",
                    "DataType":"String",
                    "_value":"Hello World"
                }
            },
            {
                "Column":{
                    "Name":"key3",
                    "DataType":"Integer",
                    "_value":"999"
                }
            }
        ]
    }
}

And for more complex XML case like this, you can check the JSON analogue here.

对于像这样更复杂的 XML 案例，您可以在此处查看 JSON 模拟。