Business
Jobs
  • About Us
  • Solutions
    • Job Postings
      Post your job and receive qualified candidates in 48h.
    • Candidate Assessments
      500+ technical and psychological tests, plus anti-fraud.
    • Headhunting
      Tailor-made executive search from start to finish.
    • Payroll + EOR
      Payroll dispersal and EOR across 15+ LATAM countries.
  • Pricing
  • Jobs

0

925
Views
How to handle null/empty values in JsonConvert.DeserializeObject

I have the following code:

return (DataTable)JsonConvert.DeserializeObject(_data, (typeof(DataTable)));

Then, I tried:

var jsonSettings = new JsonSerializerSettings
{
    NullValueHandling = NullValueHandling.Ignore
};

return (DataTable)JsonConvert.DeserializeObject<DataTable>(_data, jsonSettings);

The return line is throwing the error:

{"Error converting value \"\" to type 'System.Double'."}

Lots of solutions online suggesting creating custom Class with nullable types but this won't work for me. I can't expect the json to be in a certain format. I have no control over the column count, column type, or column names.

over 4 years ago · Santiago Trujillo
3 answers
Answer question

0

You can supply settings to JsonConvert.DeserializeObject to tell it how to handle null values, in this case, and much more:

var settings = new JsonSerializerSettings
                    {
                        NullValueHandling = NullValueHandling.Ignore,
                        MissingMemberHandling = MissingMemberHandling.Ignore
                    };
var jsonModel = JsonConvert.DeserializeObject<Customer>(jsonString, settings);
over 4 years ago · Santiago Trujillo Report

0

An alternative solution for Thomas Hagström, which is my prefered, is to use the property attribute on the member variables.

For example when we invoke an API, it may or may not return the error message, so we can set the NullValueHandling property for ErrorMessage:


    public class Response
    {
        public string Status;

        public string ErrorCode;

        [JsonProperty(NullValueHandling = NullValueHandling.Ignore)]
        public string ErrorMessage;
    }


    var response = JsonConvert.DeserializeObject<Response>(data);

The benefit of this is to isolate the data definition (what) and deserialization (use), the deserilazation needn’t to care about the data property, so that two persons can work together, and the deserialize statement will be clean and simple.

over 4 years ago · Santiago Trujillo Report

0

You can subscribe to the 'Error' event and ignore the serialization error(s) as required.

    static void Main(string[] args)
    {
        var a = JsonConvert.DeserializeObject<DataTable>("-- JSON STRING --", new JsonSerializerSettings
        {
            Error = HandleDeserializationError
        });
    }

    public static void HandleDeserializationError(object sender, ErrorEventArgs errorArgs)
    {
        var currentError = errorArgs.ErrorContext.Error.Message;
        errorArgs.ErrorContext.Handled = true;
    }
over 4 years ago · Santiago Trujillo Report
Answer question
Find remote jobs

Discover the new way to find a job!

Top jobs
Top job categories
Business
Post vacancy Pricing Sales
Legal
Terms and conditions Privacy policy
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Show me some job opportunities
There's an error!