JSON 到 XML 类的转换

Question

我似乎无法将 Json 的类排成 Linq XML。

c.first、c.second 和 c.third 被突出显示并声明：

“您是否缺少 using 指令或程序集引用。”

var serializer = new JavaScriptSerializer();
var json1 = "[count:[place:{first:1,second:2,third:3}],[place:{first:11,second:22,third:33}],[place:{first:111,second:222,third:333}]]]";
var jsons = serializer.Serialize(json1);
var jsona = serializer.Deserialize<List<jClass>>(jsons);
var xmld = new XDocument(
    new XElement("count", jsona.Select(c =>
        new XElement("place",
            new XElement("first", c.first),
            new XElement("second", c.second),
            new XElement("third", c.third)
        )
    ))
);

类.cs

public class jClass
{
    public jNumber[] count { get; set; }
}
public class jNumber
{
    public jTopThree[] place { get; set; }
}
public class jTopThree
{
    public int first { get; set; }
    public int second { get; set; }
    public int third { get; set; }
}

Answer 1

你的问题是你的对象结构本质上是一个数组数组，你只做一个选择。在您构建 xml 的位置，您的 c 变量位于 jClass 级别，因此您尝试从中读取 first、second 和 third 属性。

目前还不清楚您的 xml 结构应该是什么，但您要么需要使用更多的 .Select 调用来进一步深入到 jTopThree 实例，要么使用 .SelectMany 将其展平，或者更改您的对象定义增加了一点。

Answer 2

json 文本和对象结构似乎无效。你可以试试这样的..

对象结构

public class jClass
{
    public jTopThree[] count { get; set; }
}
public class jTopThree
{
    public int first { get; set; }
    public int second { get; set; }
    public int third { get; set; }
}

序列化

        var serializer = new JavaScriptSerializer();
        var json1 = "{count:[{first:1,second:2,third:3},{first:11,second:22,third:33},{first:111,second:222,third:333}]}";
        var obj = serializer.Deserialize<jClass>(json1);
        XmlSerializer xmlSerializer = new XmlSerializer(typeof(jClass));
        string xmlString;
        using (System.IO.StringWriter sww = new System.IO.StringWriter())
        {
            using (XmlWriter writer = XmlWriter.Create(sww))
            {
                xmlSerializer.Serialize(writer, obj);
                xmlString = sww.ToString();
            }
        }

Answer 3

        var json1 = "[{\"place\":{\"first\":1,\"second\":2,\"third\":3}},{\"place\":{\"first\":11,\"second\":22,\"third\":33}},{\"place\":{\"first\":111,\"second\":222,\"third\":333}}]";
        XmlDocument deserializeXmlNode;
        if (json1.IndexOf("[", StringComparison.OrdinalIgnoreCase) == 0)
        {
            deserializeXmlNode = Newtonsoft.Json.JsonConvert.DeserializeXmlNode($"{{\"items\":{json1}}}", "source");
        }
        else
        {
            deserializeXmlNode = Newtonsoft.Json.JsonConvert.DeserializeXmlNode(json1, "source");
        }

        Console.WriteLine(deserializeXmlNode.OuterXml);

如果您只想复制到 XML 并负担得起 Nugget 包，请使用 Newtonsoft.Json 并修改上面的代码以满足您的需求。我稍微更改了输入 json，因为它对 http://www.json.org/ 无效。不过，它应该适用于任何有效的 JSON。

http://www.newtonsoft.com/json/help/html/ConvertingJSONandXML.htm

How to convert JSON to XML or XML to JSON?

Answer 4

首先，您的 JSON 格式错误，根据您的示例，我认为它应该看起来像这样（对我来说，“地方”似乎是多余的，但我现在已经离开了）：

{
“计数”：[
{“地点”：[
{“第一”：1，“第二”：2，“第三”：3}]}，
{“地点”：[
{"first":11,"second":22,"third":33}]},
{“地点”：[
{"first":111,"second":222,"third":333}]}
]
}

你的类应该是这样的：

public class RootObject
{
    public List<Count> count { get; set; }
}

public class Count
{
    public List<Place> place { get; set; }
}

public class Place
{
    public int first { get; set; }
    public int second { get; set; }
    public int third { get; set; }
}

最后你可以用这种方式创建 XML：

var serializer = new JavaScriptSerializer();
var json = "your json string";
var result= serializer.Deserialize<RootObject>(json);
var xmld = new XDocument(
           new XElement("count", result.count
               .SelectMany(p => p.place)
               .Select(x => 
                   new XElement("place",
                       new XElement("first", x.first),
                       new XElement("second", x.second),
                       new XElement("third", x.third)
                   )
                ))
           );

输出：

<count>
  <place>
    <first>1</first>
    <second>2</second>
    <third>3</third>
  </place>
  <place>
    <first>11</first>
    <second>22</second>
    <third>33</third>
  </place>
  <place>
    <first>111</first>
    <second>222</second>
    <third>333</third>
  </place>
</count>