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如何编写按名字查找的查询

graphql c#

Andrew · 技术社区 · 5 年前

我正在尝试理解图ql,并有一个基本的示例在工作。例如,如果我传递这个查询,我将得到与ID匹配的结果。

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

我的数据只是一组静态的测试数据,现在我想做的是返回所有名字与我提供的名字匹配的用户。我很困惑怎么写这个,因为ID空支票似乎阻止了我!?

我的个人查询如下:

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"));
            });

    }
}

我该如何做才能返回名为的人员列表,不知道这是否是下面的有效查询,但希望获得一些帮助,了解如何与工作ID示例一起执行此操作。

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

回答更新-由Davidg提供

我按上面提到的做了,所以我的个人查询现在看起来像这样

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);
                });

        }
    }

然后我可以按如下方式运行graphql查询:

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

1 回复 | 直到 5 年前

DavidG 5 年前

你要么改变你在这里的领域,使 id 参数可选或创建新字段(可以调用 persons 或 people )并添加一个新的参数,将其解析到数据存储库中。就我个人而言,我更愿意做后者,并创造一个新的领域。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"));
        });
}

现在你只需要写 GetByFirstName 自己动手。现在查询如下所示:

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

现在你可能会发现 获得第一名 还不够,您还需要一个姓氏参数,并且这些参数是可选的,因此您可以这样做:

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"));
    });