Converting from JSON to XML using c#

Question

i have kind of wired requirement for converting JSON to xml. we have an API that returns the JSON response as below.

{
   "status":"Error",
   "errorMessages":{
      "1001":"Schema validation Error"
   }
}

We want to convert this JSON to XML as below using c#

<root>
  <status>ERROR</status>
  <errorMessages>
    <ErrorCode>1001</ErrorCode>
    <ErrorDescription>Schema validation Error</ErrorDescription>    
  </errorMessages>
</root>

The API team is very resistant to change the way there are generating the JSON. So i have to find a way to convert this json to XML.

I am getting the below error when i try to convert

XmlDocument doc = JsonConvert.DeserializeXmlNode(json);

"JSON root object has multiple properties. The root object must have a single property in order to create a valid XML document. Consider specifying a DeserializeRootElementName. Path errorMessages

thanks for the help in advance. :)

when i try to run the deserializeXmlNode method i am getting the below error XmlDocument doc = JsonConvert.DeserializeXmlNode(json); ` i am getting the below error when i try to convert "JSON root object has multiple properties. The root object must have a single property in order to create a valid XML document. Consider specifying a DeserializeRootElementName. Path 'errorMessages ` — abhi pm
– abhi pm, Commented Mar 21, 2018 at 12:44
You need to write a custom transformer for this JSON because 1) it has no root node (as the error says) and 2) it has properties that begin with a number (which would be invalid XML element names). There is no direct conversion path. — Dan Wilson
– Dan Wilson, Commented Mar 21, 2018 at 13:19

mu88 · Accepted Answer · 2018-03-21 14:58:49Z

2

At first, I would hardly recommend that your API team provides a valid JSON object. Otherwise you will have to write a converter that does the job. The converter could look like this:

using System.Collections.Generic;

namespace Json
{
    using System.IO;
    using System.Xml.Serialization;

    using Newtonsoft.Json.Linq;

    class Program
    {
        static void Main(string[] args)
        {
            var converter = new Converter();
            converter.Convert();
        }
    }

    class Converter
    {
        public void Convert()
        {
            //  your JSON string goes here
            var jsonString = @"{""status"":""Error"",""errorMessages"":{ ""1001"":""Schema validation Error"", ""1953"":""Another error""}}";

            // deconstruct the JSON
            var jObject = JObject.Parse(jsonString);
            var root = new Root { Status = jObject["status"].ToString(), ErrorMessages = new List<ErrorMessage>() };
            foreach (var errorMessageJsonObject in jObject["errorMessages"])
            {
                var jProperty = (JProperty)errorMessageJsonObject;
                var errorCode = System.Convert.ToInt16(jProperty.Name);
                var errorDescription = jProperty.Value.ToString();
                var errorMessage = new ErrorMessage() { ErrorCode = errorCode, ErrorDescription = errorDescription};

                root.ErrorMessages.Add(errorMessage);
            }

            // serialize as XML
            var xmlSerializer = new XmlSerializer(typeof(Root));
            string xml;
            using (StringWriter textWriter = new StringWriter())
            {
                xmlSerializer.Serialize(textWriter, root);
                xml = textWriter.ToString();
            }
        }
    }

    public class Root
    {
        public string Status;

        public List<ErrorMessage> ErrorMessages;
    }

    public class ErrorMessage
    {
        public int ErrorCode;

        public string ErrorDescription;
    }
}

With this, you will read the JSON, deconstruct it into a proper object and serialize it as XML.

answered Mar 21, 2018 at 14:58

mu88

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2 Comments

abhi pm Over a year ago

Thanks a lot @mu88 this worked. :) we have asked out API team to change it but we are not sure if that will be doing it ..

mu88 Over a year ago

Great! If you like this post, please give it your vote and/or mark it as an answer.

arslanaybars · Accepted Answer · 2018-03-21 12:37:09Z

1

if you are using asp.net web api. Its already can return xml response just add an accept header like

Accept: application/xml

answered Mar 21, 2018 at 12:37

arslanaybars

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3 Comments

abhi pm Over a year ago

when i add the above header its will fail as the xml element cannot start with an integer. <root> <status>ERROR</status> <errorMessages> <1274>Schema validation Error</1274> </errorMessages> </root>

abhi pm Over a year ago

i am consuming this service in biztalk server. So i need to write a custom c# component that does this conversion.

arslanaybars Over a year ago

check it : stackoverflow.com/questions/814001/…

Matthias Grün · Accepted Answer · 2018-03-21 12:38:12Z

1

From the documentation at https://www.newtonsoft.com/json/help/html/ConvertingJSONandXML.htm

string json = @"{
  '?xml': {
    '@version': '1.0',
    '@standalone': 'no'
  },
  'root': {
    'person': [
      {
        '@id': '1',
        'name': 'Alan',
        'url': 'http://www.google.com'
      },
      {
        '@id': '2',
        'name': 'Louis',
        'url': 'http://www.yahoo.com'
      }
    ]
  }
}";

XmlDocument doc = (XmlDocument)JsonConvert.DeserializeXmlNode(json);

answered Mar 21, 2018 at 12:38

Matthias Grün

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Comments

Carlos Henrique · Accepted Answer · 2018-03-21 12:36:55Z

0

Use the JsonConvert class which contains helper methods for this precise purpose:

// To convert an XML node contained in string xml into a JSON string   
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);
string jsonText = JsonConvert.SerializeXmlNode(doc);

// To convert JSON text contained in string json into an XML node
XmlDocument doc = JsonConvert.DeserializeXmlNode(json);

The Completed Documentation : Here

answered Mar 21, 2018 at 12:36

Carlos Henrique

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2 Comments

abhi pm Over a year ago

hi carlos thanks for the response. i am getting the below error when i try to convert "JSON root object has multiple properties. The root object must have a single property in order to create a valid XML document. Consider specifying a DeserializeRootElementName. Path 'errorMessages'"

Carlos Henrique Over a year ago

Hi Abhi pm, try it if possible for you. link

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