显示-父类型
人-子类型
一个节目可以有很多人。我想将来自shown类型的每个'personId‘作为参数传递给它的子类型Person,如下所示:
{
shows
{
showId
personId
person(personId: <I_NEED_TO_PASS_THE_personId_ABOVE_HERE>)
{
name
}
}
}我可以知道如何访问Person字段中的PersonId字段(兄弟)的值吗?
public class ShowType : ObjectGraphType<Core.Show>
{
public ShowType(
IPersonRepository PersonRepository,
IHttpContextAccessor accessor)
{
Field(a => a.ShowId);
Field(a => a.PersonId); // PERSONID_VALUE
Field<PersonType>(
"Person",
arguments: new QueryArguments(
new QueryArgument<IdGraphType> { Name = "PersonId" }
),
resolve: context =>
{
var PersonId = context.GetArgument<string>("PersonId");
return PersonRepository.GetPerson(accessor.HttpContext, <I_NEED_PERSONID_VALUE_DIRECTLY_HERE_OR_THROUGH_ARGUMENTS>);
}
);
}
}请建议如何实现这一点。
发布于 2019-10-10 16:55:00
这个问题已经解决了。多亏了SUNMO - How to access arguments in nested fields in ASP.NET Core GraphQL
public class ShowType : ObjectGraphType<Core.Show>
{
public ShowType(
IPersonRepository PersonRepository,
IHttpContextAccessor accessor)
{
Field(a => a.ShowId);
Field(a => a.PersonId);
Field<PersonType>(
"Person",
resolve: context =>
{
return PersonRepository.GetPerson(accessor.HttpContext, context.Source.PersonId);
}
);
}
}https://stackoverflow.com/questions/58288839
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