如何在 .NET 的 GraphQL 中改变自定义对象列表 [英] How to mutate a list of custom objects in GraphQL for .NET

问题描述

使用 GraphQL for .NET，我想用一个新的集合.

Using GraphQL for .NET, I would like to replace the collection of Foo with a new collection.

鉴于此服务器端代码:

public class Foo
{
    public Foo(string name)
    {
        Name = name;
    }

    public string Name { get; set; }
}

public class Root
{
    public Foo[] Foos { get; private set; }

    public Foo[] UpdateFoos(Foo[] foos)
    {
        Foos = foos;
        return Foos;
    }
}

public class MutationSchema : Schema
{
    public MutationSchema()
    {
        Query = new MutationQuery();
        Mutation = new MutationChange();
    }
}

public class FooType : ObjectGraphType
{
    public FooType()
    {
        Name = "IndividualFoo";
        Field<StringGraphType>("name");
    }
}

public class FoosType : ObjectGraphType<ListGraphType<FooType>>
{
    public FoosType()
    {
        Name = "ListOfFoo";
        Field<ListGraphType<FooType>>("foos");
    }
}

public class FoosInput : InputObjectGraphType
{
    public FoosInput()
    {
        Name = "InputForManyFoo";
        Field<ListGraphType<FooInput>>("foos");
        Field<ListGraphType<FooType>>("foosResult");
    }
}

public class FooInput : InputObjectGraphType
{
    public FooInput()
    {
        Name = "InputForSingleFoo";
        Field<StringGraphType>("name");
    }
}

public class MutationQuery : ObjectGraphType
{
    public MutationQuery()
    {
        Name = "Query";
        Field<FoosType>("queryAllFoos");
    }
}

public class MutationChange : ObjectGraphType
{
    public MutationChange()
    {
        Name = "Mutation";

        Field<FoosInput>(
            "updateAllFoos",
            arguments: new QueryArguments(
                new QueryArgument<FoosInput>
                {
                    Name = "updateFoosQueryArgument"
                }
            ),
            resolve: context =>
            {
                var root = context.Source as Root;
                var change = context.GetArgument<Foo[]>("updateFoosQueryArgument");
                // TODO: update collection e.g. return root.UpdateFoos(change);
                return change;
            }
        );
    }
}

当我运行突变查询时:

mutation M {
    fooCollection: updateAllFoos(updateFoosQueryArgument: {
        foos: [
            {name: "First Foo"},
            {name: "Second Foo"}
        ]}) {
        foosResult
    }
}

然后我收到以下错误:

{GraphQL.Validation.ValidationError: Cannot query field "foosResult" on type "InputForManyFoo". Did you mean "foosResult"?}

我正在使用 最新版本的 .NET GraphQL在撰写本文时.

I'm using the latest version of GraphQL for .NET at the time of writing.

我错过了什么?

工作示例: 如何在 GraphQL 中为 .网络

推荐答案

来自 Gitter 的回答.

My answer from Gitter.

为结果创建一个 ObjectGraphType.请注意，resolve 返回的对象的形状"与图形类型的形状"匹配.

Make an ObjectGraphType for the result. Notice that the "shape" of the object that is returned from resolve matches the "shape" of the graph type.

public class FoosResultType : ObjectGraphType
{
    public FoosResultType()
    {
        Field<ListGraphType<FooType>>("foosResult");
    }
}

public class FoosResult
{
    public IEnumerable<Foo> FoosResult { get;set; }
}

public class MutationChange : ObjectGraphType
{
    public MutationChange()
    {
        Name = "Mutation";

        Field<FoosResultType>(
            "updateAllFoos",
            arguments: new QueryArguments(
                new QueryArgument<ListGraphType<FooInput>>
                {
                    Name = "updateFoosQueryArgument"
                }
            ),
            resolve: context =>
            {
                var root = context.Source as Root;
                var change = context.GetArgument<List<Foo>>("updateFoosQueryArgument");
                // TODO: update collection e.g. return root.UpdateFoos(change);
                return new FoosResult { FoosResult = change };
            }
        );
    }
}

和更新的突变:

mutation M {
    fooCollection: updateAllFoos(updateFoosQueryArgument: [
          {name: "First Foo"},
          {name: "Second Foo"}
        ]) {
        foosResult {
          name
        }
    }
}

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