Cloud Firestore deep get with subcollection

Question

Let\'s say we have a root collection named \'todos\'.

Every document in this collection has:

1. title: String
2. subcollection named

Answer 1

As pointed out in other answers, you cannot request deep queries.

My recommendation: Duplicate your data as minimally as possible.

I'm running into this same problem with "pet ownership". In my search results, I need to display each pet a user owns, but I also need to be able to search for pets on their own. I ended up duplicated the data. I'm going to have a pets array property on each user AS WELL AS a pets subcollection. I think that's the best we can do with these kinds of scenarios.

Answer 2

you could try something like this

db.collection('coll').doc('doc').collection('subcoll').doc('subdoc') 

Hope this helps !

Answer 3

I used AngularFirestore (afs) and Typescript:

import { map, flatMap } from 'rxjs/operators';
import { combineLatest } from 'rxjs';

interface DocWithId {
  id: string;
}

convertSnapshots<T>(snaps) {
  return <T[]>snaps.map(snap => {
    return {
      id: snap.payload.doc.id,
      ...snap.payload.doc.data()
    };
  });
}

getDocumentsWithSubcollection<T extends DocWithId>(
    collection: string,
    subCollection: string
  ) {
    return this.afs
      .collection(collection)
      .snapshotChanges()
      .pipe(
        map(this.convertSnapshots),
        map((documents: T[]) =>
          documents.map(document => {
            return this.afs
             .collection(`${collection}/${document.id}/${subCollection}`)
              .snapshotChanges()
              .pipe(
                map(this.convertSnapshots),
                map(subdocuments =>
                  Object.assign(document, { [subCollection]: subdocuments })
                )
              );
          })
        ),
        flatMap(combined => combineLatest(combined))
      );
  }
  

Answer 4

This type of query isn't supported, although it is something we may consider in the future.

Answer 5

I have faced the same issue but with IOS, any way if i get your question and if you use auto-ID for to-do collection document its will be easy if your store the document ID as afield with the title field in my case :

let ref = self.db.collection("collectionName").document()

let data  = ["docID": ref.documentID,"title" :"some title"]

So when you retrieve lets say an array of to-do's and when click on any item you can navigate so easy by the path

ref = db.collection("docID/\(todo_items)")

I wish i could give you the exact code but i'm not familiar with Javascript

Answer 6

If anyone is still interested in knowing how to do deep query in firestore, here's a version of cloud function getAllTodos that I've come up with, that returns all the 'todos' which has 'todo_items' subcollection.

exports.getAllTodos = function (req, res) {
    getTodos().
        then((todos) => {
            console.log("All Todos " + todos) // All Todos with its todo_items sub collection.
            return res.json(todos);
        })
        .catch((err) => {
            console.log('Error getting documents', err);
            return res.status(500).json({ message: "Error getting the all Todos" + err });
        });
}

function getTodos(){
    var todosRef = db.collection('todos');

    return todosRef.get()
        .then((snapshot) => {
            let todos = [];
            return Promise.all(
                snapshot.docs.map(doc => {  
                        let todo = {};                
                        todo.id = doc.id;
                        todo.todo = doc.data(); // will have 'todo.title'
                        var todoItemsPromise = getTodoItemsById(todo.id);
                        return todoItemsPromise.then((todoItems) => {                    
                                todo.todo_items = todoItems;
                                todos.push(todo);         
                                return todos;                  
                            }) 
                })
            )
            .then(todos => {
                return todos.length > 0 ? todos[todos.length - 1] : [];
            })

        })
}


function getTodoItemsById(id){
    var todoItemsRef = db.collection('todos').doc(id).collection('todo_items');
    let todo_items = [];
    return todoItemsRef.get()
        .then(snapshot => {
            snapshot.forEach(item => {
                let todo_item = {};
                todo_item.id = item.id;
                todo_item.todo_item = item.data(); // will have 'todo_item.title' and 'todo_item.completed'             
                todo_items.push(todo_item);
            })
            return todo_items;
        })
}