Question

我正在使用AngularFire + FireStore开展一些项目。

我的Firestore模型如下所示：

Collection 1
- key1 {some data}
- key2 {some data}
收集2
- key1 {other data}
- key2 {其他数据}

现在我需要显示列表，其中我有来自第一个集合的一些数据，以及来自另一个集合的一些数据。如何创建这样的列表？

制作两个可观察对象然后合并它似乎毫无意义。

const collection = this.afStore.collection<any>(collectionName);
    return collection.snapshotChanges()
      .map(participants => {
        return participants.map(participant => {
          const data = participant.payload.doc.data();
          const id = participant.payload.doc.id;
          return {id, ...data};
        });
      });

我有这个带有数据+ id的代码，现在我需要使用这个id从另一个集合中提取数据，但不知道如何。

Answer 1

以下是我的做法......

“projects”数据包含customer_doc_id，customer_builder_doc_id，customer_contractor_doc_id，它们是Firestore中其他数据集合的doc_id。命令mergeMap适用于rxjs ^ 5.5.0。

 private getProjectsCombineObservables(): Observable<any[]> {
this.projectsCollection = this.afs.collection<Project>('projects', ref =>
  ref.orderBy('job_number', 'desc'));
return this.projectsCollection.snapshotChanges()
  .map(actions => {
    return actions.map(a => {
      const project_data = a.payload.doc.data() as Project;
      const doc_id = a.payload.doc.id;

      let observable1 = this.getCustomer(project_data.customer_doc_id);
      let observable2 = this.getBuilder(project_data.customer_builder_doc_id);
      let observable3 = this.getContractor(project_data.customer_contractor_doc_id);

      const combinedData = Observable.combineLatest(observable1, observable2, observable3, (data1, data2, data3) => {
        return { ...data1, ...data2, ...data3 };
      });

      return combinedData.map(data => Object.assign({}, { doc_id, ...project_data, ...data }));
    });
  }).mergeMap(observables => Observable.combineLatest(observables));

}

Answer 2

Wohoo！我做到了：）

所以它应该是这样的：

＆＃13;

const collection = this.afStore.collection<any>('collection1');
    return collection.snapshotChanges()
      .map(participants => {
        return participants.map(participant => {
          const data = participant.payload.doc.data();
          const id = participant.payload.doc.id;
          return this.afStore.doc('collection2/' + id).valueChanges()
            .map(data2 => Object.assign({}, {id, ...data, ...data2}));
        });
      }).flatMap(observables => Observable.combineLatest(observables));

＆＃13;

Answer 3

const collection = this.afStore.collection<any>('collection1');
    return collection.snapshotChanges()
      .map(participants => {
        return participants.map(participant => {
          const data = participant.payload.doc.data();
          const id = participant.payload.doc.id;
          return this.afStore.doc('collection2/' + id).valueChanges()
            .map(data2 => Object.assign({}, {id, ...data, ...data2}));
        });
      }).flatMap(observables => Observable.combineLatest(observables));

Answer 4

在尝试多种解决方案后，我使用 RXJS combineLatest, take 运算符完成了它。使用 map 函数我们可以组合结果。

combineLatest(
    this.fireBaseService.getFromCollection1(),
    this.fireBaseService.getFromCollection2(),
    //In collection 2 we have document with reference id of collection 1
)
.pipe(
    take(1),
).subscribe(
    ([dataFromCollection1, dataFromCollection2]) => {

        this.dataofCollection1 = dataFromCollection1.map((data) => {
            return {
                id: data.payload.doc.id,
                ...data.payload.doc.data() as {},
            }
            as IdataFromCollection1;
        });

        this.dataofCollection2 = dataFromCollection2.map((data2) => {
            return {
                id: data2.payload.doc.id,
                ...data2.payload.doc.data() as {},
            }
            as IdataFromCollection2;
        });

        console.log(this.dataofCollection2, 'all feeess');

        const mergeDataFromCollection =
            this.dataofCollection1.map(itm => ({
                payment: [this.dataofCollection2.find((item) => (item.RefId === itm.id))],
                ...itm
            }))

        console.log(mergeDataFromCollection, 'all data');
    },

AngularFire - 将两个FireStore集合与相同的pushId项组合在一起

4 个答案: