HotChocolate (GraphQL) 在没有查询调用的情况下不可能包含嵌套对象

Question

// Parent entity
public class Parent
{
   public int Id {get; set;}
   public int CurrencyId {get; set;}
   public virtual Currency Currency {get; set;}
}
// Currency Entity
public class Currency
{
  public int Id {get; set;}
  public string Name {get; set;}
  public virtual ICollection<Parent> Parents {get; set;}
}

// Query
[UseDbContext(typeof(AppDbContext))]
[UseFirstOrDefault]
[UseProjection]
[UseFiltering]
[UseSorting]
public IQueryable<Parent> GetParents([ScopedService] AppDbContext context, int id)
{
  return context.Set<Parent>().Where(x => x.id == id);
}

public class ParentType : ObjectType<Parent>
{
   protected override void Configure(IObjectTypeDescriptor<Parent> descriptor)
    {
        descriptor.Field("currencyName")
            .ResolveWith<Resolvers>(t => t.GetCurrencyName(default!));
    }

    private class Resolvers
    {
        public string GetCurrencyName([Parent] Parent parent)
        {
            return parent?.Currency?.Name;
        }
    }
}

当我使用带货币对象的 graphql 查询调用它时。

    query{
     parents(id: 1){
      id
      currencyName
      currency{
       name
      }
     }
    }

//Result
{
  "data": {
    "parents": {
      "id": 1,
      "currency": {
        "name": "USD"
      },
      "currencyName": "USD"
    }
  }
}

结果货币名称不为空。当我调用它时没有货币对象。

       query{
         parents(id: 1){
          id
          currencyName
         }
        }
// Result
{
  "data": {
    "parents": {
      "Id": 1,
      "currencyName": null
    }
  }
}

结果 currencyName 为空。不可能从代码中包含像货币这样的嵌套对象？我想在 graphql 查询中不调用货币对象来获取货币名称。

Answer 1

不，这目前是不可能的，因为您访问的属性只有在选择嵌套 class 时才会解析

HotChocolate (GraphQL) 在没有查询调用的情况下不可能包含嵌套对象

HotChocolate (GraphQL) it is impossible include nested object without query calling

c#

graphql

hotchocolate