Question

我似乎无法将Json的类排成Linq XML。

c.first，c.second和c.third会突出显示并说明：

“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)
        )
    ))
);

Class.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Select。在您构建xml的位置，c变量位于jClass级别，因此您尝试阅读first，second和third那个属性。

目前还不清楚你的xml结构应该是什么，但你要么需要再使用几个.Select调用来深入研究jTopThree实例，使用.SelectMany将其展平，或稍微更改您的对象定义。

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

        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并且可以买得起Nugget包，请转到Newtonsoft.Json并修改上面的代码以满足您的需求。我有slighlty更改了输入json，因为它对http://www.json.org/无效。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，你的JSON格式很糟糕，基于你的例子，我觉得它看起来应该是这样的（对我来说，＃34;地点＆＃34;似乎多余，但我现在已经离开了它）：

{
  ＆＃34;计算＆＃34;：[
  {＆＃34;地点＆＃34;：[
  {＆＃34;第一＆＃34;：1，＆＃34;第二＆＃34;：2＆＃34;第三＆＃34;：3}]}，
  {＆＃34;地点＆＃34;：[
  {＆＃34;第一＆＃34;：11，＆＃34;第二＆＃34;：22，＆＃34;第三＆＃34; 33}]}，
  {＆＃34;地点＆＃34;：[
  {＆＃34;第一＆＃34;：111，＆＃34;第二＆＃34;：222，＆＃34;第三＆＃34; 333}]}
  ]
  }

您的课程应如下所示：

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，您可以通过以下方式创建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>

JSON到XML类转换

4 个答案: