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How to Deserialize Json string to Type without creating class

By : lawtonb
Date : August 01 2020, 02:00 PM
around this issue I have an Json string.I would like to know how to Deserialize "Type" from Json string "without creating class".Please check my below code and advise how to do this.... , you can use 'dynamic' type
code :
        string MParam = @"[{'ColCode': 'BK'}]";
        dynamic result = JsonConvert.DeserializeObject<dynamic>(MParam);
        var ColCode = result[0].ColCode;

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Deserialize Json String in .Net without creating Type or anonymous type

By : jabbyjim
Date : March 29 2020, 07:55 AM
it fixes the issue You can do this easily using Json.NET.
Either parse your string as a JObject and use it like a dictionary:
code :
var obj = JObject.Parse(str);
var action = obj["Request"]["Header"]["Action"];
dynamic obj = JsonConvert.DeserializeObject<dynamic>(str);
var action = obj.Request.Header.Action;

VB.net deserialize, JSON Conversion from type 'Dictionary(Of String,Object)' to type 'String'

By : user3536681
Date : March 29 2020, 07:55 AM
hope this fix your issue The JSON provided only has one element, so it will result in a collection (dictionary) of one. I added an "item" to be sure the code below worked and for illustration purposes. Proper indentation makes things easier to follow:
code :
    "foo": {
        "HashTag": "12342345636",
        "companyname": "my test company",
        "LeadDetail": {
            "id": "1",
            "firstname": "ziggy",
            "lastname": "clark",
            "email": "emak@mai.com",
            "phone": "9874534444"
    "bar": {
        "HashTag": "02342345636",
        "companyname": "my test company2",
        "LeadDetail": {
            "id": "1",
            "firstname": "john",
            "lastname": "clark",
            "email": "emak@mai.com",
            "phone": "1874534444"
' modified from VS's EDIT -> Paste Special -> JSON as Classes
Public Class Item           
    Public Property HashTag As String
    Public Property companyname As String
    Public Property LeadDetail As Leaddetail
End Class

Public Class Leaddetail
    Public Property id As String
    Public Property firstname As String
    Public Property lastname As String
    Public Property email As String
    Public Property phone As String
End Class
Dim jstr As String = ...
' use the Item/Value class not the container
Dim myJ = JsonConvert.DeserializeObject(Of Dictionary(Of String, Item))(jstr)

' print some of the data
For Each kvp As KeyValuePair(Of String, Item) In myJ
    Console.WriteLine("key {0}, CompName {1}, Contact: {2}", kvp.Key,

Creating a class in JSON.NET to deserialize a given JSON

By : ziopiero
Date : March 29 2020, 07:55 AM
should help you out I need an example how of to build a class for a JSON response with JSON.NET. , Try using this class structure:
code :
class ResponseCall
    public ResponseHeader ResponseHeader { get; set; }

    public Response Response { get; set; }

    public Dictionary<string, object> Highlighting { get; set; }

class ResponseHeader
    public int Status { get; set; }

    public int QTime { get; set; }

    public Dictionary<string, string> Params { get; set; }

class Response
    public int NumFound { get; set; }

    public int Start { get; set; }

    public List<Dictionary<string, string>> Docs { get; set; }
ResponseCall rc = JsonConvert.DeserializeObject<ResponseCall>(json);

VB.NET Deserialize JSON - Cannot deserialize the current JSON object into type .Customer[]' because the type requires a

By : user1927184
Date : March 29 2020, 07:55 AM
Any of those help You are getting this exception because you have omitted the Property keyword from a couple of the members in your classes, which changes the semantics of the parentheses.
For example, in your CustomerWrapper class, you have declared the customers member like this:
code :
Public customers() As Customer
Public customers As Customer()   ' Field - array of Customer
Public Property customers() As Customer   ' Property - single Customer
Public Class CustomerWrapper
    Public Property customers As Customer
End Class

Public Class Metadata
    Public Property id As String
End Class

Public Class Customer
    Public Property id As String
    Public Property created_at As String
    Public Property email As String
    Public Property given_name As String
    Public Property family_name As String
    Public Property address_line1 As String
    Public Property address_line2 As String
    Public Property address_line3 As String
    Public Property city As String
    Public Property region As String
    Public Property postal_code As String
    Public Property country_code As String
    Public Property language As String
    Public Property phone_number As String
    Public Property metadata As Metadata
End Class

Deserialize without creating wrapper class using json.net

By : Ryan
Date : March 29 2020, 07:55 AM
like below fixes the issue As others have mentioned, your JSON is not valid.
Suppose the valid JSON is: { "Album": { "Name": "Classical", "Date": "2005-4-7T00:00:00" } }
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