I'm making a note app and I have a problem. It is how to check if Firebase Firestore user data collection is empty.
Query query = firebaseFirestore.collection("mNotes").document(firebaseUser.getUid()).collection("userNotes").orderBy("title", Query.Direction.ASCENDING);
FirestoreRecyclerOptions<firebasemodel> allusernotes = new FirestoreRecyclerOptions.Builder<firebasemodel>().setQuery(query, firebasemodel.class).build(); ```
this code is used to get data from Firestore.
how can I check user data collection is empty with the if statement.
I am a beginner.
How to check Firebase Firestore user data collection is empty?
Firestore doesn't have the concept of an empty collection. If a collection doesn't contain any documents, it doesn't exist at all. So there is no API that can help you check if a collection actually exists. A collection in Firestore will start to exist if there is at least one document present in it.
So to solve this, you might consider checking the number of documents within the collection or subcollection like this:
val db = Firebase.firestore
val usersRef = db.collection("users")
usersRef.get().addOnCompleteListener {
if (it.isSuccessful) {
val numberOfDocs = it.result.documents.size
if (numberOfDocs > 0) {
Log.d(TAG,"users collection already exists!")
}
}
}
I'm trying to query to my Firebase collection using the value inside of Array. I'm using arrayContains and then passing the value like shown in the code, but after querying, it returns a length of 0.
FirebaseFirestore.instance
.collection('chatrooms')
.where('users.0', arrayContains: {'studentid':'1WINXTQdshhn4jLfhMWWaZNNdL32'})
.get();
Here's an image of my database.
Database Collection
This cannot be done using array-contains. You must use the entire student object to query using array-containing. This query tries to find {'studentid':'1WINXTQdshhn4jLfhMWWaZNNdL32'} object inside the array of users.0 field.
Read more about this here https://firebase.google.com/docs/firestore/query-data/queries#array_membership.
I am using Firebase function and Firebase Firestore to develope an API which will store users data.
I wanted to locate the documents using the properties stored in their field. This is the Firebase document which states how to achieve the same.
// Create a reference to the cities collection
var citiesRef = db.collection('cities');
// Create a query against the collection
var queryRef = citiesRef.where('state', '==', 'CA');
I wanted to handle two situations
Where there is no document with the present conditions
Where there are more than two documents with the present conditions
How could the above two situation be handled?
Following our "discussion" in the comments above, in a Cloud Function you could do as follows, using the QuerySnapshot returned by the get() method:
admin.firestore().collection("cities")
.where('state', '==', 'CA')
.get()
.then(querySnapshot => {
if (querySnapshot.size == 0) {
console.log("0 documents");
} else if (querySnapshot.size > 2) {
console.log("More than 2 documents");
}
});
As said, above, just be aware that this will cost a read for each document in the collection. In case you have a very large collection, you could write a Cloud Function that update a counter each time a doc is added/removed to/from the collection.
The accepted answer does not show how to extract the data from each document and imo is only half the answer. the following will get you iterating through every document and extracting the data.
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());
});
});
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.
Like upper question, i want to get value of some field in firebase firestore instead of all document with DocumentSnapshot
like this in SQL SELECT col_1, col_2, col_3 FROM table_name
How can i do it?
Thank you for the help.
The Cloud Firestore client-side SDKs always read and returns full documents. There is no way to read a subset of the fields in a document.
You can retrieve the entire document, and then process the DocumentSnapshot to just use the fields you're interested. But this means you're using more bandwidth than needed. If this is a regular occurrence for your app, consider creating a secondary collection where each document contains just the fields you're interested in.
Also see Doug's answer here (for Swift): How to access a specific field from Cloud FireStore Firebase in Swift
Firestore can read single field value.
Firebase guides looks like didn't show these simple method.
For example:
[android]
String value = document.getString("col_1");
[node.js]
const value = doc.data().col_1
Firebase allows a few ways to query like an RDBMS database. And these are very handy too. Try the Where clause. It returns a single document but u can add multiple filters.
check for more
try this code:
//set a global variable
dataList: Array<any>;
const dataCollection = firebase.firestore().collection('data');
dataCollection.get(query => {
query.forEach(docs => {
this.dataList.push({id: doc.id, data:doc.data()});
})
})