[英]How to query firestore() for graphQL resolver?
我将GraphQL应用程序与现有的Firebase项目结合在一起,并且在获取查询以从firestore()正确获取数据时遇到很多问题。
到目前为止,我的变体工作正常,但是当我查询数据时,我无法将firestore()。get()快照转换为graphQL可以识别的形式。
到目前为止,它看起来像这样:
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-> Rooms下,我试图获取一个graphQL查询以返回“ room”集合中的所有房间。 我已经能够使用以下命令将它发送到console.log()文档列表:
return Room.get()
.then(snapshot => {
snapshot.forEach(doc => {
console.log(doc.id, " => ", doc.data());
但是到目前为止,使我难以理解的是。 任何帮助都非常感谢。
看到没人能回答这个问题,我最终自己弄清楚了:p
因此,解决与获取相关数据(例如职位)相关的功能。 以下作品:
首先,您需要一个函数将快照转换为数组,这是graphQL所期望的。 这也允许您分隔id并将其与数组项一起添加:
const snapshotToArray = (snapshot) => {
var returnArr = [];
snapshot.forEach((childSnapshot)=> {
var item = childSnapshot.data();
item.id = childSnapshot.id;
returnArr.push(item);
});
return returnArr;
};
接下来,当获取数据时,您可以使用.get()返回一个promise(和错误),该promise可以传递给snapshotToArray()。
return Position.get().then((snapshot) => {
return snapshotToArray(snapshot);
})
对于仅调用一个数据集的解决函数,例如inRoom。 它与第一个类似,除了在快照函数中使用.where()并将id和data()分开之外:
return Room.doc(parent.inRoomId).get().then((snapshot) => {
var item = snapshot.data();
item.id = snapshot.id;
return item;
})
万一有人遇到相同的问题:)
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