How to correctly assign properties for other properties in JObject?

What I am trying to achieve is converting a JObject to XML Document and then extract the outer xml from the XMl Document. Reasons behind this is to send the results as a push notification through Azure Notification Hub.

What I am trying to get is:

<toast>
    <visual>
        <binding template="ToastGeneric">
            <text id="1">Message</text>
        </binding>
    </visual>
</toast>

What I have tried:

JObject notificationPayload = new JObject(
    new JProperty("toast",
        new JObject(
            new JProperty("visual",
                new JObject(
                    new JProperty("binding",
                        new JObject(
                            new JProperty("@template", "ToastGeneric"),
                            new JProperty("text", notificationMessage,
                                new JProperty("@id", "1")))))))));

The above code is throwing an exception: Can not add Newtonsoft.Json.Linq.JProperty to Newtonsoft.Json.Linq.JArray. So what I have tried after is:

JObject notificationPayload = new JObject(
    new JProperty("toast",
        new JObject(
            new JProperty("visual",
                new JObject(
                    new JProperty("binding",
                        new JObject(
                            new JProperty("@template", "ToastGeneric"),
                            new JProperty("text", notificationMessage,
                                new JObject(
                                    new JProperty("@id", "1"))))))))));

Above code gave me a result, but not the intended one. What I got was:

<toast>
    <visual>
        <binding template="ToastGeneric">
            <text>Message</text>
            <text id="1" />
        </binding>
    </visual>
</toast>

To extract Xml From the JObject I am using the following method:

string jsonStringToConvertToXmlString = JsonConvert.SerializeObject(notificationPayload);
XmlDocument doc = JsonConvert.DeserializeXmlNode(jsonStringToConvertToXmlString);
return doc.OuterXml;

Question: How could I give the id property to the same Text property?

Jon Skeet
people
quotationmark

Basically, don't use a tool oriented at JSON to construct XML. If you already had JSON, it would make sense to use Json.NET to convert it to XML - but as you're building it from scratch, it's much cleaner to use LINQ to XML:

XDocument doc = new XDocument(
    new XElement("toast",
        new XElement("visual",
            new XElement("binding",
                new XAttribute("template", "ToastGeneric"),
                new XElement("text",
                    new XAttribute("id", 1),
                    "Message"
                )
            )
        )
    )
);

people

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