current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Dismiss
Announcing Stack Overflow Documentation

We started with Q&A. Technical documentation is next, and we need your help.

Whether you're a beginner or an experienced developer, you can contribute.

Sign up and start helping → Learn more about Documentation →
up vote 1 down vote favorite

I'm using JSON.NET Linq to JSON, when I execute the below I get the following argument exception:

Can not add Newtonsoft.Json.Linq.JValue to Newtonsoft.Json.Linq.JObject

What's wrong with the code

string content = new JObject(
            new JObject("auth",
                        new JProperty("user", "[email protected]"),
                        new JProperty("secret", "somepassword")
                ),
            new JObject("config",
                        new JProperty("template", "1")
                ),
            new JObject("data",
                        new JProperty("email", "body")
                )
            ).ToString();
share|improve this question
    
Tim is right. What you wrote looks more like the creation of an XML structure using XDocument and XElement where you can mix the node name and its child nodes in the parameter list. – fero Oct 29 '12 at 12:48
up vote 1 down vote accepted

As per the JSON.NET documentation, you need to wrap your JObjects in JProperties themselves: 


string content = new JObject(
                   new JProperty("auth",
                     new JObject(
                       new JProperty("user", "[email protected]"),
                       new JProperty("secret", "somepassword")
                     )
                   ),
                   new JProperty("config",
                     new JObject(
                       new JProperty("template", "1")
                     )
                   ),
                   new JProperty("data",
                     new JObject(
                       new JProperty("email", "body")
                     )
                   )
                 ).ToString();

share|improve this answer
    
thank you very much, simple when you know how. – Simon Oct 29 '12 at 18:48
    
Works perfectly :) – Simon Oct 29 '12 at 18:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.