How to force System.Text.Json serializer throw exception when property is missing?

user_921054 0
Viewed 5980

Json.NET behaviour could be defined by attributes: either use default or just throw an exception if json payload does not contain required property.

Yet System.Text.Json serializer silently does nothing.
Having class:

public sealed class Foo
{
    [Required]
    public int Prop {get;set;} = 10;
}

and deserializing empty object:

JsonSerializer.Deserialize<Foo>("{}");

I simply get an instance of Foo with Prop=10. I could not find any setting in JsonSerializerOptions to force it throw an exception. Is it possible?

system.text.json .net-core json c#
3 Answers
user_65270490

System.Text.Json isn't a full replacement for Json.NET. It's built for speed with minimal allocations in mind, not feature completeness. If you want validations you can

  1. Use Json.NET
  2. Validate the objects after serialisation with the Validator class
  3. Create a custom converter

TheGeneral showed how to do #3. A custom validator would have to handle all validations explicitly and return some meaningful exception though. Throwing an ArgumentNullException is enough if there's only a single property to check. Validating multiple properties would require something more complex like a ValidationException to include the validation results.

K. Scott Allen's article Manual Validation with Data Annotations shows how to do #2.

One option is to use the Validator.ValidateObject to validate an object and get a ValidationException with all the failing validations :

try
{
    var validationCtx=new ValidationContexgt(myFoo);
    Validator.ValidateObject(myFoo,validationCtx);
}
catch(ValidatinException ex)
{
    //Do something with the results.
}

This is OK if invalid objects are rare, as throwing exceptions is expensive. Perhaps a better option is to use Validator.TryValidateObject :

var results = new List<ValidationResult>();
var validationCtx=new ValidationContexgt(myFoo);
if(Validator.TryValidateObject(myFoo,validationCtx,results))
{
    //Use the object
}
else
{
    //Do something with the failed results
}
user_1342040

All you need to do SetMissingMemberHandling and it will handle every thing for you but you need to install DevBetter.JsonExtensions MissingMemberHandling.Ignore and MissingMemberHandling.Error

var deserializeOptions = new JsonSerializerOptions()
    .SetMissingMemberHandling(MissingMemberHandling.Ignore);

var weatherForecast = JsonSerializer.Deserialize<WeatherForecast>(jsonString, deserializeOptions);
user_43039440
Related