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.
How to: If I want to retrieve a FieldValue (String) only if in the same document there is another field (boolean) which is true.
For Example: If I have a Document in Firestore which contains a Field: {name: "Some Name", active: true} and I only want the name to be retrieved if the field "active" is true. Is this possible?
Does this question make sense? Sorry, I'm a self-taught developer and I have difficulties to explain stuff as a coder. But I hope you guys can still understand what I want to achieve.
You can run query to filter out the documents you don't want. like:
query(collection(db, "users"), where("active", "==", true));
But if you want retrieve a single fieldvalue like "name" and filter out other value in same document like "active" is impossible. firestore only retrieve whole document.
I am have almost this structure of data in my firestore and
I am fetching data using two queries orderBy and where but there two are not working and giving me error
the piece of code i am using is this
useEffect(() => {
db.collection("accounts")
.doc("usama_nj#gmail.com")
.collection("glasses")
.where("stock", ">", 0)
.orderBy("id", "desc")
.onSnapshot((snapshot) =>
setGlasses(snapshot.docs.map((doc) => doc.data()))
);
}, []);
if i use only single query orderBy() or where() then the query works but the combination of there query dont work together.
Solution of the problem i want is to get the data whose stock is greater then zero and also order the data according to id
You need to use .orderBy("stock").orderBy("id", "desc") to fix that query.
However this won't give the results you want. Instead you'll need to add a boolean field "hasStock" that has the same meaning as what you are querying.
I have a mongoose find method like this,
photo.find({},{
name:1,
src:1,
likes:{$literal:[]},
dislikes:{$literal:[]},
}).then(photos => ....)
what I want is, when I run the code likes and dislikes field must be an empty array for every record.
I try this way but not working.
Unsupported projection option: likes: { $literal: 1 }
Any idea to add default value for any field in find method ?
As per mongoose, document schema is constructed during its creation. So you can also edit the schema with a default value, so for every record, when it is created it will create likes, and dislikes with empty values.
You can also do this way, if you feel the schema control is not in your hands.
https://mongoosejs.com/docs/2.7.x/docs/defaults.html
photo.find({ 'name' : '1', 'likes': {$ne: []}})
I am trying to pull all the documents in the collection 'users', but it only pulls 'fred' and 'lisa', and ignores all the italicized documents:
For this data:
Trying to get all documents:
Will yield:
info: length 2
info: fred => { gender: 'male', contacts: [ '' ] }
lisa => { contacts: [ '' ] }
According to the Firebase documentation (Firebase: Add and Manage Data):
Warning: Even though non-existent ancestor documents appear in the console, they do not appear in queries and snapshots. You must create the document to include it in query results.
Note: The non-existent ancestor users seem to be auto-created when the user hits the sign-up button that triggers a firebase.auth() function (fred and lisa were created manually).
How would I print the contacts of each user if some of the users do not show up in my queries? Would I need to periodically run a script that manually re-adds all the users or is there a more elegant solution?
As you have mentioned, these "documents" are displayed with an italic font in the Firebase console: this is because these documents are only present (in the console) as "container" of one or more sub-collection but they are not "genuine" documents.
As matter of fact, if you create a document directly under a col1 collection with the full path doc1/subCol1/subDoc1, no intermediate documents will be created (i.e. no doc1 document).
The Firebase console shows this kind of "container" (or "placeholder") in italics in order to "materialize" the hierarchy and allow you to navigate to the subDoc1 document but doc1 document doesn't exist in the Firestore database.
Let's take an example: Imagine a doc1 document under the col1 collection
col1/doc1/
and another one subDoc1 under the subCol1 (sub-)collection
col1/doc1/subCol1/subDoc1
Actually, from a technical perspective, they are not at all relating to each other. They just share a part of their path but nothing else. One side effect of this is that if you delete a document, its sub-collection(s) still exist.
So, if you want to be able to query for these parent documents, you will have to create them yourself, as jackz314 mentioned in the comments.
If you're trying to list all your registered users from Firebase auth, you can use the Firebase SDK function:
function listAllUsers(nextPageToken) {
admin.auth().listUsers(1000, nextPageToken)
.then(function(listUsersResult){
listUsersResult.users.forEach(function(userRecord) {
console.log('user', userRecord.toJSON());
})
if (listUsersResult.pageToken) {
// list next batch of users
}
})
.catch(function(err) {
console.log('Error listing users: ', error)
});
}
listAllUsers();
via http://firebase.google.com/docs/auth/admin/manage-users#list_all_users