如何查询 firestore() 以获取 graphQL 解析器? [英] How to query firestore() for graphQL resolver?

问题描述

我正在将 GraphQL 应用与我现有的 Firebase 项目相结合，但在获取查询以正确从 firestore() 获取数据时遇到了很多问题.

I'm combining a GraphQL app with my existing Firebase project and am having a lot of problems getting the queries to correctly get data from the firestore().

到目前为止，我的突变工作正常，但是当我去查询数据时，我无法将 firestore().get() 快照转换为 graphQL 可以识别的形式.

So far I have the mutations working correctly, but when I go to query the data I can't get the firestore().get() snapshot into a form that graphQL will recognize.

目前看起来是这样的:

const {GraphQLObjectType,
  GraphQLString,
  GraphQLBoolean,
  GraphQLFloat,
  GraphQLSchema,
  GraphQLID,
  GraphQLList,
  GraphQLNonNull} = require("graphql");
const admin = require('firebase-admin');
const functions = require('firebase-functions');


admin.initializeApp(functions.config().firebase);

//Models
const Room = admin.firestore().collection('room');
const Position = admin.firestore().collection('position');
const Plant = admin.firestore().collection('plant');
const PlantInfo = admin.firestore().collection('plantInfo');

const RoomType = new GraphQLObjectType({
  name: "Room",
  fields: () => ({
    id: { type: GraphQLID },
    name: { type: GraphQLString },
    description: { type: GraphQLString },
    floor: { type: GraphQLString },
    building: { type: GraphQLString },
    positions: {
      type: new GraphQLList(PositionType),
      resolve(parent, arg) {
        //return _.filter(positions, {inRoomId:parent.id})
        return Position.orderByChild('inRoomId').equalTo(parent.id);
      }
    }
  })
});

const PositionType = new GraphQLObjectType({
  name: "Position",
  fields: () => ({
    id: { type: GraphQLID },
    name: { type: GraphQLString },
    description: { type: GraphQLString },
    exposure: { type: GraphQLString },
    size: { type: GraphQLString },
    inRoom: {
      type: RoomType,
      resolve(parent, args) {
        //return _.find(rooms, {id:parent.inRoomId})
        return Room.child(parent.inRoomId);
      }
    }
  })
});

const RootQuery = new GraphQLObjectType({
  name: "RootQueryType",
  fields: {
    room: {
      type: RoomType,
      args: { id: { type: GraphQLID } },
      resolve(parent, args) {
        //code to get data from db/othersourse
        //return _.find(rooms, {id: args.id});
        return Room.child(args.id);
      }
    },
    position: {
      type: PositionType,
      args: { id: { type: GraphQLID } },
      resolve(parent, args) {
        //code to get data from db/othersourse
        //return _.find(positions, {id: args.id})
        return Position.child(args.id);
      }
    },
    rooms: {
      type: new GraphQLList(RoomType),
      resolve(parent, args) {
        //return rooms
        return Room.get().then(snapshot => {snapshot.forEach(doc => {return doc})})

      }
    },
    positions: {
      type: new GraphQLList(PositionType),
      resolve(parent, args) {
        //return positions
        return Position.get().then(doc => console.log(doc)).catch(err => console.log('Error getting document', err));
      }
    }
  }
});

const Mutation = new GraphQLObjectType({
  name: "Mutation",
  fields: {
    addRoom: {
      type: RoomType,
      args: {
        name: { type: new GraphQLNonNull(GraphQLString) },
        floor: { type: new GraphQLNonNull(GraphQLString) },
        building: { type: new GraphQLNonNull(GraphQLString) }
      },
      resolve(parent, args) {
        let room = {
          name: args.name,
          floor: args.floor,
          building: args.building
        };
        return Room.add(room);
      }
    },
    addPosition: {
      type: PositionType,
      args: {
        name: { type: new GraphQLNonNull(GraphQLString) },
        exposure: { type: new GraphQLNonNull(GraphQLString) },
        size: { type: new GraphQLNonNull(GraphQLString) },
        inRoomId: { type: new GraphQLNonNull(GraphQLString) }
      },
      resolve(parent, args) {
        let position = {
          name: args.name,
          exposure: args.exposure,
          size: args.size,
          inRoomId: args.inRoomId
        };
        return Position.add(position);
      }
    }
  }
});

module.exports = new GraphQLSchema({
  query: RootQuery,
  mutation: Mutation
});

在 RootQuery -> 房间下，我试图获取一个 graphQL 查询以返回我的房间"集合中的所有房间.我已经能够使用以下命令将其发送到 console.log() 文档列表:

Under the RootQuery -> Rooms I'm trying to get a graphQL query to return all the rooms in my 'room' collection. I have been able to get it to console.log() a list of documents using:

return Room.get()
.then(snapshot => {
snapshot.forEach(doc => {
        console.log(doc.id, " => ", doc.data());

但到目前为止，我一直无法将其放入数组中.任何帮助都非常感谢.

But getting this into an array has so far eluded me. Any help is really appreciated.

推荐答案

看到没有人能够回答这个问题，我最终自己解决了 :p

Seeing as no one was able to answer this, I ended up figuring it out for myself :p

因此解决与获取相关数据集合(例如位置)相关的函数.以下作品:

So resolve functions relating to getting a collection of related data for example positions. the following works:

首先，您需要一个函数将快照转换为数组，因为这是 graphQL 所期望的.这也允许您分离 id 并将其添加到数组项中:

first you need a function to convert the snapshots into an array as this is what graphQL is expecting. This also allows your to seperate the id and add it in with the array item:

const snapshotToArray = (snapshot) => {
  var returnArr = [];

  snapshot.forEach((childSnapshot)=> {
      var item = childSnapshot.data();
      item.id = childSnapshot.id;

      returnArr.push(item);
  });

  return returnArr;
};

接下来在获取数据时，您使用 .get() 返回一个承诺(和错误)，该承诺(和错误)可以传递到 snapshotToArray() 中.

Next when getting the data you use .get() which returns a promise (and error) which can be passed into the snapshotToArray().

return Position.get().then((snapshot) => {
          return snapshotToArray(snapshot);
        }) 

对于仅调用一个数据集的解析函数，例如 inRoom.除了在快照函数中使用 .where() 和分隔 id 和 data() 之外，它与第一个类似:

For resolve functions that only call on one dataset for example inRoom. Its similar to the first one except using .where() and seperating the id and data() in the snapshot functions:

return Room.doc(parent.inRoomId).get().then((snapshot) => {
          var item = snapshot.data();
          item.id = snapshot.id;
          return item;
        })

以防万一其他人遇到同样的问题:)

Just incase someone else runs into the same problem :)

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