GraphQL中的多对多关系

Question

假设我有两个对象数组：

let movies = [
    { id: '1', title: 'Erin Brockovich'},
    { id: '2', title: 'A Good Year'},
    { id: '3', title: 'A Beautiful Mind'},
    { id: '4', title: 'Gladiator'}
];

let actors = [
    { id: 'a', name: 'Julia Roberts'},
    { id: 'b', name: 'Albert Finney'},
    { id: 'c', name: 'Russell Crowe'}
];

我想在他们之间建立多对多关系。首先使用Vanilla JavaScript，最后使用GraphQL模式。

In JavaScript I did something like this:

let movies = [
    { id: '1', title: 'Erin Brockovich',  actorId: ['a', 'b'] },
    { id: '2', title: 'A Good Year',      actorId: ['b', 'c'] },
    { id: '3', title: 'A Beautiful Mind', actorId: ['c'] },
    { id: '4', title: 'Gladiator',        actorId: ['c'] }
];
let actors = [
    { id: 'a', name: 'Julia Roberts', movieId: ['1'] },
    { id: 'b', name: 'Albert Finney', movieId: ['1', '2'] },
    { id: 'c', name: 'Russell Crowe', movieId: ['2', '3', '4'] }
];
let actorIds = [];
let movieIds = [];
for (let m = 0; m < movies.length; m ++) {
    for (let i = 0; i < movies[m].actorId.length; i ++) {
        actorIds.push(movies[m].actorId[i]);
    }
}
for (let a = 0; a < actors.length; a ++) {
    for (let i = 0; i < actors[a].movieId.length; i ++) {
        movieIds.push(actors[a].movieId[i]);
    }
}
for (let a = 0; a < actors.length; a ++) {
    for (let i = 0; i < actorIds.length; i ++) {
        if ((actors[a].id == 'c') && (actors[a].id == actorIds[i])) {
            for (let j = 0; j < movies.length; j ++) {
                if (movies[j].id == movieIds[i]) {
                    console.log(movies[j].title);
                }
            }
        }
    }
}

When I run preceding code in Node, Terminal will return

A Good Year
A Beautiful Mind
Gladiator

and this is exactly what I want.

Unfortunately I've get lost in GraphQL schema. What I have so far—inside of fields function of course—is this:

in_which_movies: {
    type: new GraphQLList(FilmType),
    resolve(parent, args) {

        let actorIds = [];
        let movieIds = [];

        for (let m = 0; m < movies.length; m ++) {
            for (let i = 0; i < movies[m].actorId.length; i ++) {
                actorIds.push(movies[m].actorId[i]);
            }
        }
        for (let a = 0; a < actors.length; a ++) {
            for (let i = 0; i < actors[a].movieId.length; i ++) {
                movieIds.push(actors[a].movieId[i]);
            }
        }
        for (var a = 0; a < actors.length; a ++) {
            for (var i = 0; i < actorIds.length; i ++) {
                if ((actors[a].id == parent.id) && (actors[a].id == actorIds[i])) {
                    for (var j = 0; j < movies.length; j ++) {
                        if (movies[j].id == movieIds[i]) {
                            console.log(movies[j].title);
                        }
                    }
                }
            }
        }
        return movies[j].title;
    }
}

When I run the following query in GraphiQL...

{
    actor(id: "c") {
        name
        in_which_movies {
            title
        }
    }
}

...I have this response:

{
  "errors": [
    {
      "message": "Cannot read property 'title' of undefined",
      "locations": [
        {
          "line": 4,
          "column": 3
        }
      ],
      "path": [
        "actor",
        "in_which_movies"
      ]
    }
  ],
  "data": {
    "actor": {
      "name": "Russell Crowe",
      "in_which_movies": null
    }
  }
}

...which is strange for me, cuz Terminal is responding as I expected

A Good Year
A Beautiful Mind
Gladiator

I guess all code I've wrote so far is useless and I need some fresh guidelines how to write many-to-many relationship in GraphQL properly.

Answer 1

TL; DR我认为您太想太多了。您的解析器做的工作太多了，导致代码难以推理和调试。

我认为您的问题与GraphQL并没有太大关系，而只是对基础数据进行正确的操作。我将尝试从您的示例以及GraphQL逐步解决它，因此我们最终提供了您正在寻找的类型和解析器。

从您的原始代码开始：

let movies = [
    { id: '1', title: 'Erin Brockovich',  actorId: ['a', 'b'] },
    { id: '2', title: 'A Good Year',      actorId: ['b', 'c'] },
    { id: '3', title: 'A Beautiful Mind', actorId: ['c'] },
    { id: '4', title: 'Gladiator',        actorId: ['c'] }
];
let actors = [
    { id: 'a', name: 'Julia Roberts', movieId: ['1'] },
    { id: 'b', name: 'Albert Finney', movieId: ['1', '2'] },
    { id: 'c', name: 'Russell Crowe', movieId: ['2', '3', '4'] }
];

我想建议我们将其翻译为索引为的内容，id以便于查询。这也将更好地为通常在生产GraphQL API之后的数据库或键值存储建模。

将其转换为已索引但仍为原始JS的内容：

let movies = {
    '1': { id: '1', title: 'Erin Brockovich',  actorId: ['a', 'b'] },
    '2': { id: '2', title: 'A Good Year',      actorId: ['b', 'c'] },
    '3': { id: '3', title: 'A Beautiful Mind', actorId: ['c'] },
    '4': { id: '4', title: 'Gladiator',        actorId: ['c'] }
};
let actors = {
    'a': { id: 'a', name: 'Julia Roberts', movieId: ['1'] },
    'b': { id: 'b', name: 'Albert Finney', movieId: ['1', '2'] },
    'c': { id: 'c', name: 'Russell Crowe', movieId: ['2', '3', '4'] }
};

接下来，我们应该考虑将代表这些类型的GraphQL模式。这是“多对多”部分起作用的地方。我认为我们可以从您的示例数据中清晰地得出类型：

type Movie {
  id: ID!
  title: String
  actors: [Actor]
}

type Actor {
  id: ID!
  name: String
  movies: [Movie]
}

请注意，这[Movie]是一个Movie对象列表。即使基础数据包含ID（也就是我们希望的那样，也称为“规范化”），我们也会根据实际的类型关系对API进行建模。

接下来，我们需要设置解析器。让我们看一下Actor类型的解析器，因为这就是您的示例。

movies: {
    type: new GraphQLList(FilmType),
    resolve(parent) {
        // The ids of all the movies this actor is in. "parent" will be the
        // actor data currently being queried
        let movieIds = parent.movieId;
        // We'll build up a list of the actual movie datas to return.
        let actorInMovies = [];
        for (let m = 0; m < movieIds.length; m++) {
            // The m'th movie id.
            let movieId = movieIds[m];
            // The movie data from our indexed "movies" top level object.
            // In production, this might access a database service
            let movie = movies[movieID];
            // Add that movie to the list of movies
            actorInMovies.push(movie)
        }
        // Then we'll return that list of movie objects.
        return actorInMovies;
    }
}

请注意，在您的原始解析程序中，您返回movies[j].title的可能是字符串，并且与“ List of FilmType”所期望的不匹配，并且在上面的示例中，我返回了电影数据对象的数组。

另外，上面的代码是执行此操作的非常冗长的方法，但是我认为对每个步骤进行注释将很有帮助。要真正实现多对多，则该Movie类型的actors字段应具有几乎相同的代码。但是，只是为了显示一个示例，说明如何通过使用.map()运算符来大大简化此代码，我将用另一种方式编写它：

actors: {
    type: new GraphQLList(ActorType),
    resolve(parent) {
        // In this case, "parent" will be the movie data currently being queried.
        // Use "map" to convert a list of actor ids into a list of actor data 
        // objects using the indexed "actors" top level object.
        return parent.actorId.map(id => actors[id]);
    }
}