如何从XML在C＃解析为JSON时避免了根元素 [英] How to avoid the ROOT Element when parsing from XML to JSON in C#

问题描述

我想创建一个XML使用JSON.NET的JSON

  XmlDocument的docPerson =新的XmlDocument（）;
docPerson.LoadXml（xmlPerson）; // xmlPerson是从上面的XML
JObject personDefinition = JObject.Parse（JsonConvert.SerializeObject（docPerson））;
Tasks.personDefinition = personDefinition [personDefinition];
 

输出JSON

 人：{人：[{PERSONID：1，名字：名1，姓氏：lastName1}，{ PERSONID：3，名字：名称2，姓氏：lastName2}]}
 

我不需要外部的人，只需要下面的JSON

必需的JSON

<$p$p><$c$c>"person":[{"personId":"1","firstName":"1","lastName":"Weight"},{"personId":"3","firstName":"4","lastName":"Pulse牛}]}

我使用的方法。首先但是它抛出一个错误都试过了。任何人都可以提出如何从输出JSON剥离根

解决方案

JSON.Net有一个 SerializeXmlNode（）方法，可以做的正是你想要的。简单地真传递给 omitRootObject 参数。 注意：

 字符串xmlPerson = @"<PersonDefinition><Person><name>Nicolas</name><age>22</age></Person><Person><name>Pankaj</name><age>25</age></Person></PersonDefinition>";
XmlDocument的docPerson =新的XmlDocument（）;
docPerson.LoadXml（xmlPerson）;
JSON字符串= JsonConvert.SerializeXmlNode（docPerson，Formatting.None，真正的）;
Console.WriteLine（JSON）;
 

输出：

  {人：[{名：尼古拉，时代：22}，{名：潘卡，时代： 25}]}
 

I am trying to create a JSON from an xml using JSON.NET

XmlDocument docPerson = new XmlDocument();
docPerson.LoadXml(xmlPerson); //xmlPerson is the xml from above
JObject personDefinition = JObject.Parse(JsonConvert.SerializeObject(docPerson));
Tasks.personDefinition = personDefinition["personDefinition"];

OUTPUT JSON

"person":{"person":[{"personId":"1","firstName":"Name1","lastName":"lastName1"},                  {"personId":"3","firstName":"Name2","lastName":"lastName2"}]}

I don't need the outer person and just need the following json

REQUIRED JSON

"person":[{"personId":"1","firstName":"1","lastName":"Weight"},{"personId":"3","firstName":"4","lastName":"Pulse Ox"}]}

I have tried using the .first method however it is throwing an error. Can anyone suggest on how to strip the root from the output json

解决方案

JSON.Net has a SerializeXmlNode() method that can do exactly what you want. Simply pass true to the omitRootObject parameter. Observe:

string xmlPerson = @"<PersonDefinition><Person><name>Nicolas</name><age>22</age></Person><Person><name>Pankaj</name><age>25</age></Person></PersonDefinition>";
XmlDocument docPerson = new XmlDocument();
docPerson.LoadXml(xmlPerson);
string json = JsonConvert.SerializeXmlNode(docPerson, Formatting.None, true);
Console.WriteLine(json);

Output:

{"Person":[{"name":"Nicolas","age":"22"},{"name":"Pankaj","age":"25"}]}

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