Related
I'm using flutter and firebase. I use pagination, max 5 documents per page. How do I know if there are more documents left to get from a firestore collection. I want to use this information to enable/disable a next page button presented to the user.
limit: 5 (5 documents each time)
orderBy: "date" (newest first)
startAfterDocument: latestDocument (just a variable that holds the latest document)
This is how I fetch the documents.
collection.limit(5).orderBy("date", descending: true).startAfterDocument(latestDocument).get()
I thought about checking if the number of docs received from firestore is equal to 5, then assume there are more docs to get. But this will not work if I there are a total of n * 5 docs in the collection.
I thought about getting the last document in the collection and store this and compare this to every doc in the batches I get, if there is a match then I know I've reach the end, but this means one excess read.
Or maybe I could keep on getting docs until I get an empty list and assume I've reached the end of the collection.
I still feel there are a much better solution to this.
Let me know if you need more info, this is my first question on this account.
There is no flag in the response to indicate there are more documents. The common solution is to request one more document than you need/display, and then use the presence of that last document as an indicator that there are more documents.
This is also what the database would have to do to include such a flag in its response, which is probably why this isn't an explicit option in the SDK.
You might also want to check the documentation on keeping a distributed count of the number of documents in a collection as that's another way to determine whether you need to enable the UI to load a next page.
here's a way to get a large data from firebase collection
let latestDoc = null; // this is to store the last doc from a query
//result
const dataArr = []; // this is to store the data getting from firestore
let loadMore = true; // this is to check if there's more data or no
const initialQuery = async () => {
const first = db
.collection("recipes-test")
.orderBy("title")
.startAfter(latestDoc || 0)
.limit(10);
const data = await first.get();
data.docs.forEach((doc) => {
// console.log("doc.data", doc.data());
dataArr.push(doc.data()); // pushing the data into the array
});
//! update latest doc
latestDoc = data.docs[data.docs.length - 1];
//! unattach event listeners if no more docs
if (data.empty) {
loadMore = false;
}
};
// running this through this function so we can actual await for the
//docs to get from firebase
const run = async () => {
// looping until we get all the docs
while (loadMore) {
console.log({ loadMore });
await initialQuery();
}
};
I have a collection in firebase cloud firestore called 'posts' and I want to show the most liked posts in the last 24h on my web app.
The post documents have a field called 'like_count' (number) and another field called 'time_posted' (timestamp).
I also want to be able to limit the results to apply pagination.
I tried to apply a filter to only get the posts posted in the last 24 hours and then ordering them by the 'like_count' and then the 'time_posted' since I want the posts with the most likes to appear first.
postsRef.where("time_posted", ">", twentyFourHoursAgo)
.orderBy("like_count", "desc")
.orderBy("time_posted", "desc")
.limit(10)
However, I quickly found out that it is not possible to filter and then sort by two different fields.
(See the Limitations part of the documentation for Order and limit data with Cloud Firestore)
It states:
Invalid: Range filter and first orderBy on different fields
I thought about sorting the results by 'like_count' in the frontend, but this won't work properly because I don't have all the documents. And getting all the documents is infeasible for a large number of daily posts.
Is there an easy work-around I am missing or how can I go about this?
When performing a query, Firestore must be able to traverse an index in a continuous fashion.
This introduction video is a little outdated (because "OR" queries are now possible using the "in" operator) but it does give a good visualization of what Firestore is doing as it runs a query.
If your query was just postsRef.orderBy("like_count", "desc").limit(10), Firestore would load up the index it has for a descending "like_count", pluck the first 10 entries and return them.
To handle your query, it would have to pluck an entry off the descending "like_count" index, compare it to your "time_posted" requirement, and either discard it or add it to a list of valid entries. Once it has all of the recent posts, it then needs to sort the results as you specified. As these steps don't make use of a continuous read of an index, it is disallowed.
The solution would be to build your own index from the recent posts and then pluck the results off of that. Because doing this on the client is ill-advised, you should instead make use of a Cloud Function to do the work for you. The following code makes use of a Callable Cloud Function.
const MS_TWENTY_FOUR_HOURS = 24 * 60 * 60 * 1000;
export getRecentTopPosts = function.https.onCall((data, context) => {
// unless otherwise stated, return only 10 entries
const limit = Number(data.limit) || 10;
const postsRef = admin.firestore().collection("posts");
// OPTIONAL CODE SEGMENT: Check Cached Index
const twentyFourHoursAgo = Date.now() - MS_TWENTY_FOUR_HOURS;
const recentPostsSnapshot = await postsRef
.where("time_posted", ">", twentyFourHoursAgo)
.get();
const orderedPosts = recentPostsSnapshot.docs
.map(postDoc => ({
snapshot: postDoc,
like_count: postDoc.get("like_count"),
time_posted: postDoc.get("time_posted")
})
.sort((p1, p2) => {
const deltaLikes = p2.like_count - p1.like_count; // descending sort based on like_count
if (deltaLikes !== 0) {
return deltaLikes;
}
return p2.time_posted - p1.time_posted; // descending sort based on time_posted
});
// OPTIONAL CODE SEGMENT: Save Cached Index
return orderedPosts
.slice(0, limit)
.map(post => ({
_id: post.snapshot.id,
...post.snapshot.data()
}));
})
If this code is expected to be called by many clients, you may wish to cache the index to save it getting constantly rebuilt by inserting the following segments into the function above.
// OPTIONAL CODE SEGMENT: Check Cached Index
if (!data.skipCache) { // allow option to bypass cache
const cachedIndexSnapshot = await admin.firestore()
.doc("_serverCache/topRecentPosts")
.get();
const oneMinuteAgo = Date.now - 60000;
// if the index was created in the past minute, reuse it
if (cachedIndexSnapshot.get("timestamp") > oneMinuteAgo) {
const recentPostMetadataArray = cachedIndexSnapshot.get("posts");
const recentPostIdArray = recentPostMetadataArray
.slice(0, limit)
.map((postMeta) => postMeta.id)
const postDocs = await fetchDocumentsWithId(postsRef, recentPostIdArray); // see https://gist.github.com/samthecodingman/aea3bc9481bbab0a7fbc72069940e527
// postDocs is not ordered, so we need to be able to find each entry by it's ID
const postDocsById = {};
for (const doc of postDocs) {
postDocsById[doc.id] = doc;
}
return recentPostIdArray
.map(id => {
// may be undefined if not found (i.e. recently deleted)
const postDoc = postDocsById[id];
if (!postDoc) {
return null; // deleted post, up to you how to handle
} else {
return {
_id: postDoc.id,
...postDoc.data()
};
}
});
}
}
// OPTIONAL CODE SEGMENT: Save Cached Index
if (!data.skipCache) { // allow option to bypass cache
await admin.firestore()
.doc("_serverCache/topRecentPosts")
.set({
timestamp: Date.now(),
posts: orderedPosts
.slice(0, 25) // cache the maximum expected amount
.map(post => ({
id: post.snapshot.id,
like_count: post.like_count,
time_posted: post.time_posted,
}))
});
}
Other improvements you could add to this function include:
A field mask - i.e. instead of return every part of the post documents, return just the title, like count, time posted and the author.
Variable post age (instead of 24 hours)
Variable minimum likes count
Filter by author
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 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!`);
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.