C# Convert JSON to XML with Attribute and Value

Question

I have a json like below

{
    "Title": null,
    "ProjectName": {
        "@Label": "Project Name",
        "Value": "Test Project Name",
        "@Template": ""
    },
    "ModelNo": {
        "@Label": "Model",
        "Value": "Test Model Number",
        "@Template": ""
    }    
}

I want to convert it to xml like below:

<Root>
    <Title>Test Title</Title>
    <ProjectName Label="Project Name" Template="">Test Project Name</ProjectName>
    <ModelNo Label="Model" Template="">Test Model Number</ModelNo>
</Root>

At present, I used something by JsonConvert.DeserializeXmlNode(Json.ToString()) and add XmlAttribute, the result will be

<Root>
    <Title>Test Title</Title>
    <ProjectName Label="Project Name" Template=""><Value>Test Project Name</Value></ProjectName>
    <ModelNo Label="Model" Template=""><Value>Test Model Number</Value></ModelNo>
</Root>

How to make sure there is no Value, and map Value to xml text

According to the docs it seems like you could try renaming the Value property to #text in your JSON. — Good Night Nerd Pride
– Good Night Nerd Pride, Commented Apr 8, 2024 at 10:31

Guru Stron · Accepted Answer · 2024-04-08 10:49:10Z

You can replace/build a new JSON via LINQ to JSON API. For example (assuming the source values are in the JSON, otherwise analyze and manipulate the jp.Value):

var js = """
         {
             "Title": "Test Title",
             "ProjectName": {
                 "@Label": "Project Name",
                 "Value": "Test Project Name",
                 "@Template": ""
             },
             "ModelNo": {
                 "@Label": "Model",
                 "Value": "Test Model",
                 "@Template": ""
             }
         }
         """;
var jObject = JObject.Parse(js);
var jsonProperties = jObject.Descendants()
    .OfType<JProperty>()
    .Where(jp => jp.Name == "Value")
    .ToList();
foreach (JProperty jp in jsonProperties)
{
  // replace "Value" with "#text" property name 
  jp.Replace(new JProperty("#text", jp.Value)); 
}

var deserializeXmlNode = JsonConvert.DeserializeXmlNode(jObject.ToString(), "Root");

Console.WriteLine(deserializeXmlNode.OuterXml);

Result (formatted):

<Root>
    <Title>Test Title</Title>
    <ProjectName Label="Project Name" Template="">Test Project Name</ProjectName>
    <ModelNo Label="Model" Template="">Test Model</ModelNo>
</Root>

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