I thought I read that you can query subcollections with the new Firebase Firestore, but I don't see any examples. For example I have my Firestore setup in the following way:
Dances [collection]
danceName
Songs [collection]
songName
How would I be able to query "Find all dances where songName == 'X'"
Update 2019-05-07
Today we released collection group queries, and these allow you to query across subcollections.
So, for example in the web SDK:
db.collectionGroup('Songs')
.where('songName', '==', 'X')
.get()
This would match documents in any collection where the last part of the collection path is 'Songs'.
Your original question was about finding dances where songName == 'X', and this still isn't possible directly, however, for each Song that matched you can load its parent.
Original answer
This is a feature which does not yet exist. It's called a "collection group query" and would allow you query all songs regardless of which dance contained them. This is something we intend to support but don't have a concrete timeline on when it's coming.
The alternative structure at this point is to make songs a top-level collection and make which dance the song is a part of a property of the song.
UPDATE
Now Firestore supports array-contains
Having these documents
{danceName: 'Danca name 1', songName: ['Title1','Title2']}
{danceName: 'Danca name 2', songName: ['Title3']}
do it this way
collection("Dances")
.where("songName", "array-contains", "Title1")
.get()...
#Nelson.b.austin Since firestore does not have that yet, I suggest you to have a flat structure, meaning:
Dances = {
danceName: 'Dance name 1',
songName_Title1: true,
songName_Title2: true,
songName_Title3: false
}
Having it in that way, you can get it done:
var songTitle = 'Title1';
var dances = db.collection("Dances");
var query = dances.where("songName_"+songTitle, "==", true);
I hope this helps.
UPDATE 2019
Firestore have released Collection Group Queries. See Gil's answer above or the official Collection Group Query Documentation
Previous Answer
As stated by Gil Gilbert, it seems as if collection group queries is currently in the works. In the mean time it is probably better to use root level collections and just link between these collection using the document UID's.
For those who don't already know, Jeff Delaney has some incredible guides and resources for anyone working with Firebase (and Angular) on AngularFirebase.
Firestore NoSQL Relational Data Modeling - Here he breaks down the basics of NoSQL and Firestore DB structuring
Advanced Data Modeling With Firestore by Example - These are more advanced techniques to keep in the back of your mind. A great read for those wanting to take their Firestore skills to the next level
What if you store songs as an object instead of as a collection? Each dance as, with songs as a field: type Object (not a collection)
{
danceName: "My Dance",
songs: {
"aNameOfASong": true,
"aNameOfAnotherSong": true,
}
}
then you could query for all dances with aNameOfASong:
db.collection('Dances')
.where('songs.aNameOfASong', '==', true)
.get()
.then(function(querySnapshot) {
querySnapshot.forEach(function(doc) {
console.log(doc.id, " => ", doc.data());
});
})
.catch(function(error) {
console.log("Error getting documents: ", error);
});
NEW UPDATE July 8, 2019:
db.collectionGroup('Songs')
.where('songName', isEqualTo:'X')
.get()
I have found a solution.
Please check this.
var museums = Firestore.instance.collectionGroup('Songs').where('songName', isEqualTo: "X");
museums.getDocuments().then((querySnapshot) {
setState(() {
songCounts= querySnapshot.documents.length.toString();
});
});
And then you can see Data, Rules, Indexes, Usage tabs in your cloud firestore from console.firebase.google.com.
Finally, you should set indexes in the indexes tab.
Fill in collection ID and some field value here.
Then Select the collection group option.
Enjoy it. Thanks
You can always search like this:-
this.key$ = new BehaviorSubject(null);
return this.key$.switchMap(key =>
this.angFirestore
.collection("dances").doc("danceName").collections("songs", ref =>
ref
.where("songName", "==", X)
)
.snapshotChanges()
.map(actions => {
if (actions.toString()) {
return actions.map(a => {
const data = a.payload.doc.data() as Dance;
const id = a.payload.doc.id;
return { id, ...data };
});
} else {
return false;
}
})
);
Query limitations
Cloud Firestore does not support the following types of queries:
Queries with range filters on different fields.
Single queries across multiple collections or subcollections. Each query runs against a single collection of documents. For more
information about how your data structure affects your queries, see
Choose a Data Structure.
Logical OR queries. In this case, you should create a separate query for each OR condition and merge the query results in your app.
Queries with a != clause. In this case, you should split the query into a greater-than query and a less-than query. For example, although
the query clause where("age", "!=", "30") is not supported, you can
get the same result set by combining two queries, one with the clause
where("age", "<", "30") and one with the clause where("age", ">", 30).
I'm working with Observables here and the AngularFire wrapper but here's how I managed to do that.
It's kind of crazy, I'm still learning about observables and I possibly overdid it. But it was a nice exercise.
Some explanation (not an RxJS expert):
songId$ is an observable that will emit ids
dance$ is an observable that reads that id and then gets only the first value.
it then queries the collectionGroup of all songs to find all instances of it.
Based on the instances it traverses to the parent Dances and get their ids.
Now that we have all the Dance ids we need to query them to get their data. But I wanted it to perform well so instead of querying one by one I batch them in buckets of 10 (the maximum angular will take for an in query.
We end up with N buckets and need to do N queries on firestore to get their values.
once we do the queries on firestore we still need to actually parse the data from that.
and finally we can merge all the query results to get a single array with all the Dances in it.
type Song = {id: string, name: string};
type Dance = {id: string, name: string, songs: Song[]};
const songId$: Observable<Song> = new Observable();
const dance$ = songId$.pipe(
take(1), // Only take 1 song name
switchMap( v =>
// Query across collectionGroup to get all instances.
this.db.collectionGroup('songs', ref =>
ref.where('id', '==', v.id)).get()
),
switchMap( v => {
// map the Song to the parent Dance, return the Dance ids
const obs: string[] = [];
v.docs.forEach(docRef => {
// We invoke parent twice to go from doc->collection->doc
obs.push(docRef.ref.parent.parent.id);
});
// Because we return an array here this one emit becomes N
return obs;
}),
// Firebase IN support up to 10 values so we partition the data to query the Dances
bufferCount(10),
mergeMap( v => { // query every partition in parallel
return this.db.collection('dances', ref => {
return ref.where( firebase.firestore.FieldPath.documentId(), 'in', v);
}).get();
}),
switchMap( v => {
// Almost there now just need to extract the data from the QuerySnapshots
const obs: Dance[] = [];
v.docs.forEach(docRef => {
obs.push({
...docRef.data(),
id: docRef.id
} as Dance);
});
return of(obs);
}),
// And finally we reduce the docs fetched into a single array.
reduce((acc, value) => acc.concat(value), []),
);
const parentDances = await dance$.toPromise();
I copy pasted my code and changed the variable names to yours, not sure if there are any errors, but it worked fine for me. Let me know if you find any errors or can suggest a better way to test it with maybe some mock firestore.
var songs = []
db.collection('Dances')
.where('songs.aNameOfASong', '==', true)
.get()
.then(function(querySnapshot) {
var songLength = querySnapshot.size
var i=0;
querySnapshot.forEach(function(doc) {
songs.push(doc.data())
i ++;
if(songLength===i){
console.log(songs
}
console.log(doc.id, " => ", doc.data());
});
})
.catch(function(error) {
console.log("Error getting documents: ", error);
});
It could be better to use a flat data structure.
The docs specify the pros and cons of different data structures on this page.
Specifically about the limitations of structures with sub-collections:
You can't easily delete subcollections, or perform compound queries across subcollections.
Contrasted with the purported advantages of a flat data structure:
Root-level collections offer the most flexibility and scalability, along with powerful querying within each collection.
I'm trying to figure out a way to get a collection of documents along with their subcollection data.
For example, my tree is...
-locations
-{locationId}
-historic
-april
-airTemp: 12.4
-displayOrder: 4
-august
-airTemp: 14.5
-displayOrder: 9
-december
-airTemp: 13.5
-displayOrder: 12
etc..
...where locationId is a document and historic is the subcollection with monthly documents in it.
I know how to get the top level collection items and store it into an array but I want to add their subcollection data (i.e. the jan, feb, etc.) into each document as well.
var locations = []
let ref = db.collection('locations')
db.collection('locations').get()
.then(snapshot => {
snapshot.forEach(doc => {
let location = doc.data()
location.id = doc.id
location.name = doc.name
// get the historic subcollection data here
})
})
How can I get the combined data (collection and subcollection) from each object and then push it into an array?
bounty is wrong structure
each month should be its own object
There is no way, with the mobile/web client libraries, to get, in one query, the data of a Firestore document and the data of its sub-collection(s) documents.
There is even no way to get the list of the sub-collections of a document (with the mobile/web client libraries), see https://firebase.google.com/docs/firestore/query-data/get-data#list_subcollections_of_a_document
So you need to know the name of the sub-collections in order to query them, which is your case (name = 'historic').
You could do as follows. First you get all the parent documents and at the same time you use Promise.all() to query, in parallel, all the 'historic' sub-collections.
var db = firebase.firestore();
var promises = []
db.collection('locations').get()
.then(snapshot => {
snapshot.forEach(doc => {
let location = doc.data()
location.id = doc.id
location.name = doc.data().name
console.log(location);
promises.push(doc.ref.collection('historic').get());
})
return Promise.all(promises);
})
.then(results => {
results.forEach(querySnapshot => {
querySnapshot.forEach(function (doc) {
console.log(doc.id, " => ", doc.data());
});
});
});
Note that the order of the results array is exactly the same than the order of the promises array.
Also note that to get the value of a document item (e.g. name) you need to do doc.data().name and not doc.name.
First create an empty array, e.g. resultsArray = []. Use simple or compound queries to get at the documents that you want to read. Then iterate over the resultDocuments creating an resultObject recording whatever properties you want. Inside that iteration, .push(resultObject) to the results array.
I'm using the firestore of firebase and I want to iterate through the whole collection. Is there something like:
db.collection('something').forEach((doc) => {
// do something
})
Yes, you can simply query the collection for all its documents using the get() method on the collection reference. A CollectionReference object subclasses Query, so you can call Query methods on it. By itself, a collection reference is essentially an unfiltered query for all of its documents.
Android: Query.get()
iOS/Swift: Query.getDocuments()
JavaScript: Query.get()
In each platform, this method is asynchronous, so you'll have to deal with the callbacks correctly.
See also the product documentation for "Get all documents in a collection".
db.collection("cities").get().then(function(querySnapshot) {
querySnapshot.forEach(function(doc) {
// doc.data() is never undefined for query doc snapshots
console.log(doc.id, " => ", doc.data());
});
});
If you know that there aren't too many docs in the collection (e.g. thousands or millions) then you can just use collectionRef.get() as described in the top-voted answer here and explained in Firebase docs.
However, in many cases, a collection can contain large numbers of documents that you can't just "get" at once, as your program's memory usage will explode. In these cases you need to implement a different traversal logic that will go through the entire collection by batches. You also need to ensure that you don’t miss any documents or process any of them multiple times.
This is why I wrote Firecode, an open-source Node.js library that solves precisely this problem. It is an extremely light, robust, well-typed, and well-documented library that provides you with configurable traverser objects that walk you through a given collection.
You can find the Github repo here and the docs site here. Also, here's a short snippet that shows you how you would traverse a users collection with Firecode.
const usersCollection = firestore().collection('users');
const traverser = createTraverser(usersCollection);
const { batchCount, docCount } = await traverser.traverse(async (batchDocs, batchIndex) => {
const batchSize = batchDocs.length;
await Promise.all(
batchDocs.map(async (doc) => {
const { email, firstName } = doc.data();
await sendEmail({ to: email, content: `Hello ${firstName}!` });
})
);
console.log(`Batch ${batchIndex} done! We emailed ${batchSize} users in this batch.`);
});
console.log(`Traversal done! We emailed ${docCount} users in ${batchCount} batches!`);
Note: Querying across subcollections is not currently supported in Cloud Firestore. If you need to query data across collections, use root-level collections.
When I read the documentation for me it's not clear if it's possible to do something like this:
var listsRef = db.collection("users/user1/lists");
var query = listsRef.where("public", "==", true);
I can understand that it's not supported if I try to do this:
var listsRef = db.collection("users/{userId}/lists");
var query = listsRef.where("public", "==", true);
But I am wondering if I can query in a specific collection that comes from a specific document. In fact, what's the difference here and in a root collection?
Thanks in advance.
The SDK now supports this
Please see the other answers below...
Old answer
You can query on a subcollection as long as you know the document it belongs to. For example the following is a valid subcollection query:
const usersCollection = firestore.collection("users");
const adminDocument = usersCollection.doc("admin");
const adminsFollowersQuery = adminDocument.collection("followers").where("name", "==", "Arthur Dent");
adminsFollowersQuery.get().then((adminsFollowers) => {
adminsFollowers.docs.forEach((adminFollower) => {
console.log(adminFollower.get("name"));
});
});
As you point out the following is NOT currently valid:
const allUserFollowersQuery = firestore.collection("users/{wildcard}/followers")
.where("name", "==", "Arthur Dent");
adminsFollowersQuery.get().then((allUserFollowers) => {
allUserFollowers.docs.forEach((follower) => {
console.log(follower.get("name"));
});
});
I think the documentation here is a little confusing but by "Querying across subcollections" they mean, you can query a specific subcollection, as above, but you cannot query general subcollections. At least not as a single action using the FireStore SDK.
Firebase now supports querying subcollections (as of May 2019).
You can do this
db.collectionGroup("lists").where("public", "==", true)
which will return matching lists documents for all users.
I'm trying to model "memberships" with Firestore. The idea is that there are companies, users and then memberships.
The memberships collection stores a reference to a company and to a user, as well as a role as a string, e..g admin or editor.
How would I query to get the users with a certain role for a company?
This is what I currently have with some basic logging.
const currentCompanyID = 'someid';
return database
.collection('memberships')
.where('company', '==', database.doc(`companies/${currentCompanyID}`))
.where('role', '==', 'admin')
.get()
.then(snap => {
snap.forEach(function(doc) {
console.log(doc.id, ' => ', doc.data());
const data = doc.data();
console.log(data.user.get());
});
})
.catch(error => {
console.error('Error fetching documents: ', error);
});
data.user.get() returns a promise to the user, but I'd have to do that for every user which seems inefficient?
What would be the best way to approach this?
Your code is close to what you want, but there are two issues:
Your where() clause can't compare a field with a document reference, because Firestore is a classic denormalized datastore. There aren't ways to strongly guarantee that one document refers to another. You'll need to store document IDs and maintain consistency yourself. (Example below).
Queries actually return a QuerySnapshot, which includes all the docs that result from a query. So you're not getting one document at a time — you'll get all the ones that match. (See code below)
So a corrected version that fits the spirit of what you want:
const currentCompanyID = '8675309';
const querySnapshot = await database
.collection('memberships')
.where('companyId', '==', currentCompanyID)
.where('role', '==', 'admin')
.get(); // <-- this promise, when awaited, pulls all matching docs
await Promise.all(querySnapshot.map(async snap => {
const data = doc.data();
const user = await database
.collection('users')
.doc(data.userId)
.get();
console.log(doc.id, ' => ', data);
console.log(user);
});
There isn't a faster way on the client side to fetch all the users that your query refers to at once -- it's part of the trouble of trying to use a denormalized store for queries that feel much more like classic relational database queries.
If this ends up being a query you run often (i.e. get users with a certain role within a specific company), you could consider storing membership information as part of the user doc instead. That way, you could query the users collection and get all the matching users in one shot.