CODEWITHSUNDEEP
c#jsondeserializationspecial-characters
aggy
aggy

Reputation: 173

Special characters in property name

In my Application a webservice will return a json

{
    "UserDetails": [
      {
        "UserId": "57",
        "EmailId": "[email protected]",
        "UserName": "Prasant",
        "TimezoneMins": "330",
        "CustomLogo": "",
        "IsExecutive": "True",
        "HasNTID": "1",
        "Org_Id": "1",
        "Org_Name": "Summit",
        "Designation": null,
        "Location": null,
        "Location_Name": "",
        "WsVersion": "V5.1.0",
        "CallMe": "FALSE",
        "GPS": "FALSE",
        "Feedback_IM&SR": "NULL",
        "RPT_Widgets_Access": "False"
      }
    ]   }

Here i want to Deserialize this json into a class object. In which the class contains the properties same as the keys in json.

public class UserDetails
    {
        int? _userId;
        string _emailId;
        string _userName;
        int _timezoneMins;
        string _customLogo;
        string _isExecutive;
        int _hasNTID;
        int? _org_Id;
        string _org_Name;
        string _designation;
        int? _location;
        string _location_Name;
        string _feedback_IMSR;
        string _rPT_Widgets_Access;
        public string Feedback_IM&SR
        {
           get{return _feedback_IMSR;}
           set{_feedback_IMSR = value;}
        }

    }

Here the variable and property "Feedback_IM&SR" is having '&' character which is not allowed either in variable and Property names,but i need the value of that property.

Can anyone help me on this please.

Upvotes: 7

Views: 20674

Answers (3)

Ranjith.V
Ranjith.V

Reputation: 326

You can make class objects from json using following references

http://json2csharp.com/

Just paste your json string and it will generate C# class and properties inside that json.

Upvotes: 5

MichaelThePotato
MichaelThePotato

Reputation: 1525

you can all each property however you want, but in the C# implelmentation write it down like this:

class UserDetails
{
   [JsonProperty ("Feedback_IM&SR")]
    public string FeedbackProperty{ get; set; }
}

If you the "JsonProperty" attribute above the property, the serializer will know that in the JSON, the property's name is the one stated in the attribute

Upvotes: 8

Nick Bull
Nick Bull

Reputation: 9846

Use Netwonsoft.NET:

var details = JsonConvert.DeserializeObject<UserDetails>(json);

For your class, you need to have attributes on your properties for the names:

[JsonProperty(PropertyName = "Feedback_IM&SR")]
string _feedback_imsr { get; set; }

Now you can keep the JSON data having whatever names it wishes to, and have your C# class have another name for the property. And as part of your class, that would look like:

public class UserDetails
{
    int? _userId;
    string _emailId;
    string _userName;
    int _timezoneMins;
    string _customLogo;
    string _isExecutive;
    int _hasNTID;
    int? _org_Id;
    string _org_Name;
    string _designation;
    int? _location;
    string _location_Name;
    string _wsVersion;
    string _callMe;
    string _gPS;
    string _feedback_IMSR;
    string _rPT_Widgets_Access;
    [JsonProperty(PropertyName = "Feedback_IM&SR")]
    public string Feedback_IMSR
    {
       get{return _feedback_IMSR;}
       set{_feedback_IMSR = value;}
    }
}

Upvotes: 20

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