Question

我试图理解Graph QL，并有一个基本的示例工作。例如如果我在此查询中传递，则会获得ID为ID的匹配项。

query {
  person(id:"4090D8F6-EFC4-42CD-B55C-E2203537380C")
  {
    firstname
    surname
  }
}

我的数据只是测试数据的静态集合，我现在想做的是返回所有名字与我提供的名字匹配的用户。我不知如何写，因为id null检查似乎阻止了我！？

我的PersonQuery看起来像这样：

public class PersonQuery : ObjectGraphType<Person>
{

    public PersonQuery(ShoppingData data)
    {
        Field<PersonType>(
            "person",
            description: "A Person",
            arguments: new QueryArguments(
                new QueryArgument<NonNullGraphType<IdGraphType>>
                {
                    Name = "id",
                    Description = "The id of the person"
                }),
            resolve: ctx =>
            {
                return data.GetById(ctx.GetArgument<Guid>("id"));
            });

    }
}

我将如何做到这一点，以便我可以按名字返回一个人的名单，不知道下面是否这是一个有效的查询，但是想要一些有关如何使用工作ID的帮助例子。

query {
  person
  {
    firstname: ("Andrew")
    surname
  }
}

答案更新-由DavidG提供

我按照提到的那样做，所以我的PersonQuery现在看起来像这样

public class PersonQuery : ObjectGraphType<Person>
    {

        public PersonQuery(ShoppingData data)
        {
            Field<PersonType>(
                name: "person",
                description: "A Person",
                arguments: new QueryArguments(
                    new QueryArgument<IdGraphType>
                    {
                        Name = "id",
                        Description = "The id of the person"
                    }),
                resolve: ctx =>
                {
                     return data.GetById(ctx.GetArgument<Guid>("id"));
                });

            Field<ListGraphType<PersonType>>(
                name : "persons",
                description: "Persons",
                arguments: new QueryArguments(
                    new QueryArgument<StringGraphType>
                    {
                        Name = "firstname",
                        Description = "The firstname of the person"
                    },
                    new QueryArgument<StringGraphType>
                    {
                        Name = "surname",
                        Description = "The surname of the person"
                    }),
                resolve: ctx =>
                {
                    var firstName = ctx.GetArgument<String>("firstname");
                    var surname = ctx.GetArgument<String>("surname");
                    return data.Filter(firstName, surname);
                });

        }
    }

然后我可以如下运行graphql查询：

query {
  persons(firstname: "Andrew", surname: "P")
  {
    firstname
    surname
  }
}

Answer 1

您需要更改此处的字段以使id参数为可选，或者创建一个新字段（可能称为persons或people）并添加一个新参数，您解析到数据存储库中。就个人而言，我更喜欢后者，并开辟了一个新领域。例如：

public PersonQuery(ShoppingData data)
{
    Field<PersonType>( /* snip */ );

    //Note this is now returning a list of persons
    Field<ListGraphType<PersonType>>(
        "people", //The new field name
        description: "A list of people",
        arguments: new QueryArguments(
            new QueryArgument<NonNullGraphType<StringGraphType>>
            {
                Name = "firstName", //The parameter to filter on first name
                Description = "The first name of the person"
            }),
        resolve: ctx =>
        {
            //You will need to write this new method
            return data.GetByFirstName(ctx.GetArgument<string>("firstName"));
        });
}

现在您只需要自己编写GetByFirstName方法即可。查询现在看起来像这样：

query {
  people(firstName:"Andrew")
  {
    firstname
    surname
  }
}

现在您可能会发现GetByFirstName是不够的，并且您还想要一个姓氏参数，并且它们是可选参数，因此您可以执行以下操作：

Field<ListGraphType<PersonType>>(
    "people",
    description: "A list of people",
    arguments: new QueryArguments(
        new QueryArgument<StringGraphType>
        {
            Name = "firstName", //The parameter to filter on first name
            Description = "The first name of the person"
        },
        new QueryArgument<StringGraphType>
        {
            Name = "surname",
            Description = "The surname of the person"
        }),
    resolve: ctx =>
    {
        //You will need to write this new method
        return data.SearchPeople(
            ctx.GetArgument<string>("firstName"), 
            ctx.GetArgument<string>("surame"));
    });

GraphQL如何编写查询以按名字查找

1 个答案: