Deserialize json object into dynamic object using Json net

Question

Is it possible to return a dynamic object from a json deserialization using json net  I would like to do something like this   dynamic jsonResponse   JsonConvert Deserialize json   Console WriteLine jsonResponse message

User · Answer

Yes you can do it using the JsonConvert DeserializeObject  To do that  just simple do   dynamic jsonResponse   JsonConvert DeserializeObject json   Console WriteLine jsonResponse  message

User · Answer

Yes it is possible  I have been doing that all the while   dynamic Obj   JsonConvert DeserializeObject  lt your json string gt      It is a bit trickier for non native type  Suppose inside your Obj  there is a ClassA  and ClassB objects  They are all converted to JObject  What you need to do is   ClassA ObjA   Obj ObjA ToObject lt ClassA gt     ClassB ObjB   Obj ObjB ToObject lt ClassB gt

User · Answer

If you just deserialize to dynamic you will get a JObject back   You can get what you want by using an ExpandoObject   var converter   new ExpandoObjectConverter        dynamic message   JsonConvert DeserializeObject lt ExpandoObject gt  jsonString  converter

User · Answer

As of Json NET 4 0 Release 1  there is native dynamic support    Test  public void DynamicDeserialization         dynamic jsonResponse   JsonConvert DeserializeObject     message     Hi           jsonResponse Works   true      Console WriteLine jsonResponse message      Hi     Console WriteLine jsonResponse Works      True     Console WriteLine JsonConvert SerializeObject jsonResponse         message   Hi   Works  true      Assert That jsonResponse  Is InstanceOf lt dynamic gt          Assert That jsonResponse  Is TypeOf lt JObject gt          And  of course  the best way to get the current version is via NuGet   Updated  11 12 2014  to address comments   This works perfectly fine  If you inspect the type in the debugger you will see that the value is  in fact  dynamic  The underlying type is a JObject  If you want to control the type  like specifying ExpandoObject  then do so

User · Answer

Note  At the time I answered this question in 2010  there was no way to deserialize without some sort of type  this allowed you to deserialize without having go define the actual class and allowed an anonymous class to be used to do the deserialization      You need to have some sort of type to deserialize to  You could do something along the lines of   var product   new   Name       Price   0    dynamic jsonResponse   JsonConvert Deserialize json  product GetType      My answer is based on a solution for  NET 4 0 s build in JSON serializer  Link to deserialize to anonymous types is here   http   blogs msdn com b alexghi archive 2008 12 22 using-anonymous-types-to-deserialize-json-data aspx

User · Answer

Json NET allows us to do this   dynamic d   JObject Parse   number 1000  str  string   array   1 2 3 4 5 6       Console WriteLine d number   Console WriteLine d str   Console WriteLine d array Count     Output    1000  string  6   Documentation here  LINQ to JSON with Json NET  See also JObject Parse and JArray Parse

User · Answer

I know this is old post but JsonConvert actually has a different method so it would be  var product   new   Name       Price   0    var jsonResponse   JsonConvert DeserializeAnonymousType json  product

User · Answer

If you use JSON NET with old version which didn t JObject   This is another simple way to make a dynamic object from JSON  https   github com chsword jdynamic  NuGet Install  PM gt  Install-Package JDynamic   Support using string index to access member like   dynamic json   new JDynamic   a  a 1      Assert AreEqual 1  json  a    a       Test Case  And you can use this util as following    Get the value directly  dynamic json   new JDynamic  1       json Value   2 Get the member in the json object  dynamic json   new JDynamic   a  abc        json a is a string  abc   dynamic json   new JDynamic   a 3 1416       json a is 3 1416m  dynamic json   new JDynamic   a 1       json a is integer  1   3 IEnumerable  dynamic json   new JDynamic   1 2 3      json Length json Count is 3   And you can use json 0   json 2  to get the elements  dynamic json   new JDynamic   a  1 2 3        json a Length  json a Count is 3    And you can use  json a 0   json a 2  to get the elements  dynamic json   new JDynamic    b 1   c 1        json Length json Count is 2    And you can use the  json 0  b json 1  c to get the num    Other  dynamic json   new JDynamic   a  a 1          json a a is 1

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