Csharp C#使用属性和值将JSON转换为XML

发布于04月08日

我有一个json像下面这样

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

我想把它转换成xml，如下所示:

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

目前，我用的东西按JsonConvert.DeserializeXmlNode(Json.ToString())加XmlAttribute，结果就是

<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>

如何确保没有Value，并将Value映射到xml文本

推荐答案

您可以通过LINQ to JSON API替换/构建新的JSON.例如(假设源值在JSON中，否则分析和操作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);

结果(格式):

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