XML 到 JSON - 将 Json 对象转换为 Json 数组

Question

在我的控制器中，我正在像这样将XML数据转换为JSON

 var doc = new XmlDocument();
 doc.Load(xmlStream);
 doc.DocumentElement.RemoveAttribute("xmlns"); // don't want xmlns node in json
 return Json(doc.DocumentElement.ChildNodes, new JsonSerializerSettings { StringEscapeHandling = StringEscapeHandling.Default }, Encoding.UTF8);

XML 如下所示

<Employees xmlns="http://company.com/schemas">
    <Employee>
        <FirstName>name1</FirstName>
        <LastName>surname1</LastName>
    <Employee>
    <Employee>
        <FirstName>name2</FirstName>
        <LastName>surname2</LastName>
    <Employee>
    <Employee>
        <FirstName>name3</FirstName>
        <LastName>surname3</LastName>
    <Employee>
</Employees>

并输出 JSON

[
    {
        "Employee": {
            "FirstName": "name1",
            "LastName": "surname1"
         }
    },
    {
        "Employee": {
            "FirstName": "name2",
            "LastName": "surname2"
         }
    },{
        "Employee": {
            "FirstName": "name3",
            "LastName": "surname3"
         }
    }
]

我想要实现的目标

[
    {
        "FirstName": "name1",
        "LastName": "surname1"
    },
    {
        "FirstName": "name2",
        "LastName": "surname2"
    },{
        "FirstName": "name3",
        "LastName": "surname3"
    }
]

那么有没有办法在将XML元素序列化为JSON的同时将其转换为JSON数组呢？我想使它动态化，所以我不想在这样做时编写实体类。

Answer 1

我离你的预期输出应该是多少。我认为我们可以做得更好，但这是我的两条线解决方案：

在线试用！

var doc = XDocument.Parse(xmlData);
var json = JsonConvert.SerializeXNode(doc);

输出：

{
  "Employees": {
    "@xmlns": "http://company.com/schemas",
    "Employee": [{
      "FirstName": "name1",
      "LastName": "surname1"
    }, {
      "FirstName": "name2",
      "LastName": "surname2"
    }, {
      "FirstName": "name3",
      "LastName": "surname3"
    }]
  }
}

从@Peter编辑：

由于模式是常量，因此可以导航对象以访问对象：

var employees = JObject.Parse(json).Children().Children().Children().ElementAt(1).Children().First();

在线试用！

Answer 2

如果您之前在对象数组中反序列化它是可能的。像这样：

    class Employee
    {
        public string FirstName { get; set; }
        public string LastName { get; set; }
    }
    static void DoXmlToJson()
    {
        string xmlData = @"<?xml version='1.0' encoding='UTF-8'?>
                          <Employees>
                            <Employee>
                                <FirstName>name1</FirstName>
                                <LastName>surname1</LastName>
                            </Employee>
                            <Employee>
                                <FirstName>name2</FirstName>
                                <LastName>surname2</LastName>
                            </Employee>
                            <Employee>
                                <FirstName>name3</FirstName>
                                <LastName>surname3</LastName>
                            </Employee>
                        </Employees>";
        XDocument doc = XDocument.Parse(xmlData);
        var array = doc.Root.Elements()
            .Select(row => new Employee() {
                FirstName = row.Element("FirstName").Value,
                LastName = row.Element("LastName").Value
            });
        string json = JsonConvert.SerializeObject(array);
        Console.WriteLine(json);
    }

使用 XmlSerializer.Deserialze 的另一种方法：

    public class Employee
    {
        public string FirstName { get; set; }
        public string LastName { get; set; }
    }
    public class Employees
    {
        [XmlElement("Employee")]
        public Employee[] Items { get; set; }
    }
    static void DoXmlToJson()
    {
        string xmlData = @"<Employees>
                            <Employee>
                                <FirstName>name1</FirstName>
                                <LastName>surname1</LastName>
                            </Employee>
                            <Employee>
                                <FirstName>name2</FirstName>
                                //<LastName>surname2</LastName>
                            </Employee>
                            <Employee>
                                <FirstName>name3</FirstName>
                                <LastName>surname3</LastName>
                            </Employee>
                        </Employees>";

        XmlSerializer xmlSerializer = new XmlSerializer(typeof(Employees));
        MemoryStream memoryStream = new MemoryStream(Encoding.UTF8.GetBytes(xmlData));
        Employees wholeObject = (Employees)xmlSerializer.Deserialize(memoryStream);
        string json = JsonConvert.SerializeObject(wholeObject.Items);
        Console.WriteLine(json);
    }

输出在这两种情况下看起来都是这样的：

[
  {
    "FirstName": "name1",
    "LastName": "surname1"
  },
  {
    "FirstName": "name2",
    "LastName": "surname2"
  },
  {
    "FirstName": "name3",
    "LastName": "surname3"
  }
]

Answer 3

使用 Cinchoo ETL - 一个开源库，您只需几行代码即可轻松地完成 Xml 到 Json

string xml = @"<Employees xmlns=""http://company.com/schemas"">
    <Employee>
        <FirstName>name1</FirstName>
        <LastName>surname1</LastName>
    </Employee>
    <Employee>
        <FirstName>name2</FirstName>
        <LastName>surname2</LastName>
    </Employee>
    <Employee>
        <FirstName>name3</FirstName>
        <LastName>surname3</LastName>
    </Employee>
</Employees>
";
StringBuilder sb = new StringBuilder();
using (var p = ChoXmlReader.LoadText(xml)
    )
{
    using (var w = new ChoJSONWriter(sb))
        w.Write(p);
}
Console.WriteLine(sb.ToString());

输出：

[
 {
  "FirstName": "name1",
  "LastName": "surname1"
 },
 {
  "FirstName": "name2",
  "LastName": "surname2"
 },
 {
  "FirstName": "name3",
  "LastName": "surname3"
 }
]

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