c# - Json 属性作为节点生成

Question

我需要将传入的 json 文件转换为 XML。

我正在使用以下代码来实现要求。我正在使用 Newtonsoft Json 转换器

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

输入文件是

{"menu": {   "id": "file",   "value": "File",   "popup": {     "menuitem": [       {"value": "New", "onclick": "CreateNewDoc()"},       {"value": "Open", "onclick": "OpenDoc()"}, {"value": "Close", "onclick": "CloseDoc()"}  ]  }}}

我得到的输出是

<menu><id>file</id><value>File</value><popup><menuitem><value>New</value><onclick>CreateNewDoc()</onclick></menuitem><menuitem><value>Open</value><onclick>OpenDoc()</onclick></menuitem><menuitem><value>Close</value><onclick>CloseDoc()</onclick></menuitem></popup></menu>

所以这里的属性是作为节点生成的。

提前致谢

Answer 1

来自 Converting between JSON and XML :

Conversion Rules

• Attributes are prefixed with an @ and should be at the start of the object.

因此你可以使用 Linq to JSON使用原始值修改所有 JSON 属性，并在它们的名称前添加一个 @ 字符。请注意，由于您的文件将“以 MB 为单位”，因此您应该避免加载到临时字符串，而是 stream the file contents in directly :

        // Load the JObject directly from a file
        using (var streamReader = File.OpenText(fileName))
        using (var jsonReader = new JsonTextReader(streamReader))
        {
            obj = JObject.Load(jsonReader);
        }

        // Rename all properties with primitive values (string, number, boolean, null) to begin with "@"
        foreach (var o in obj.Descendants().OfType<JObject>())
        {
            // Attributes must appear first in the JObject's property list.
            int insertIndex = 0;
            foreach (var property in o.Properties().Where((p => p.Value is JValue && !p.Name.StartsWith("@"))).ToList())
            {
                property.Remove();
                ((IList<JToken>)o).Insert(insertIndex++, new JProperty("@" + property.Name, property.Value));
            }
        }

        // Convert to XmlDocument
        XmlDocument doc;
        using (var reader = obj.CreateReader())
        {
            doc = (XmlDocument)JsonExtensions.DeserializeXmlNode(reader);
        }

使用辅助方法:

public static class JsonExtensions
{
    public static XmlDocument DeserializeXmlNode(JsonReader reader)
    {
        return DeserializeXmlNode(reader, null, false);
    }

    public static XmlDocument DeserializeXmlNode(JsonReader reader, string deserializeRootElementName, bool writeArrayAttribute)
    {
        var converter = new Newtonsoft.Json.Converters.XmlNodeConverter() { DeserializeRootElementName = deserializeRootElementName, WriteArrayAttribute = writeArrayAttribute };
        var jsonSerializer = JsonSerializer.CreateDefault(new JsonSerializerSettings { Converters = new JsonConverter[] { converter } });
        return (XmlDocument)jsonSerializer.Deserialize(reader, typeof(XmlDocument));
    }
}

这会产生输出:

<menu id="file" value="File">
  <popup>
    <menuitem value="New" onclick="CreateNewDoc()" />
    <menuitem value="Open" onclick="OpenDoc()" />
    <menuitem value="Close" onclick="CloseDoc()" />
  </popup>
</menu>