Is there a way to ignore all fields by default on a GraphQL type and only add the wanted field?

user_11023824 0
Viewed 1976

Is there a way to ignore all fields by default on a GraphQL type and only add the wanted field?

Hot Chocolate infers GraphQL type members form the C# type automatically.

This means that the following code ...

public class Foo
{
    public string Bar { get; set; }

    public string? Baz { get; set; }
}

public class FooType : ObjectType<Foo>
{
}

will result in the following GraphQL type:

type Foo {
  bar: String!
  baz: String
}

In my use-case I want to change this behavior and define explicitly which type member of my C# type is used in the GraphQL type.

hotchocolate graphql asp.net .net
1 Answers

Hot Chocolate allows you to reverse the behavior per type or for the whole schema.

To declare all fields on one specific type, explicitly do the following:

public class FooType : ObjectType<Foo>
{
    protected override void Configure(IObjectTypeDescriptor<Person> descriptor)
    {
         // this defines that fields shall only be defined explicitly
         descriptor.BindFieldsExplicitly();

         // now declare the fields that you want to define.
         descriptor.Field(t => t.Bar);    
    }
}

type Foo {
  bar: String!
}

If you want to declare fields explicitly on all types in your schema, you can set the following option:

services
    .AddGraphQLServer()
    .AddQueryType<Query>()
    // this option will, by default, define that you want to declare everything explicitly.
    .ModifyOptions(c => c.DefaultBindingBehavior = BindingBehavior.Explicit);

If you set it globally, you can override it always per type, meaning you can define in that case to bind members implicitly on a by type basis.

user_110238240
Related