Say I have this kind of structure
A (collection): {
a (doc): {
name:'Tim',
B (collection):{
b (doc): {
color:'blue'
}
}
}
}
where A and B are collections while a and b are documents.
Is there a way to get everything contained in a root document with one query?
If I query like this
db.collection("A").doc("a").get()
I just gets name:'Tim' field. What I want is to also get all B's documents.
I basically wish my query returns
{
user:'Tim',
B (collection):{
b (doc): {
color:'blue'
}
}
}
Is it possibly or do I really need to make multiple queries one for each collection :/ ?
Say I have a really deep nested tree of collections representing the user profile, my costs will raise like hell since each time I load a user profile I have a multiplier of read requests 1 x N where N is the depth of my tree :/.
If you are concerned about costs of each pull, you will need to structure your data according to your common view / pull needs, rather than what you might prefer for a perfect structure. If you need to pull these things together every time, Consider using "maps" for things that do not actually need to be sub-collections with documents.
In this example, "preferences" is a map.
{
user: "Tim",
preferences: {
color: "blue",
nickname: "Timster"
}
}
Each document is also limited in size to 1MB - so if you need to store something for this user that will scale and continue to grow, like log records, then it would make sense to break logs into a sub-collection that only gets pulled when you want it, making each log entry a separate document... And whether all logs for all users are stored in a separate parent collection, or a sub-collection of each user really depends on how you will be pulling logs and what will result in fast speeds, balanced against costs of pulls. If you're showing this user their last 10 searches, then a search-log would make good sense as a sub-collection. If you're pulling all search data for all users for analysis, then a separate parent level collection would make sense because you can pull all logs in 1 pull, to prevent the need to pull logs from each user separately.
You can also nest your pulls and promises together for convenience purposes.
// Get reference to all of the documents
console.log("Retrieving list of documents in collection");
let documents = collectionRef.limit(1).get()
.then(snapshot => {
snapshot.forEach(doc => {
console.log("Parent Document ID: ", doc.id);
let subCollectionDocs = collectionRef.doc(doc.id).collection("subCollection").get()
.then(snapshot => {
snapshot.forEach(doc => {
console.log("Sub Document ID: ", doc.id);
})
}).catch(err => {
console.log("Error getting sub-collection documents", err);
})
});
}).catch(err => {
console.log("Error getting documents", err);
});
As we know querying in Cloud Firestore is shallow by default. This type of query isn't supported, although it is something Google may consider in the future.
Adding to Matt R answer, if you're using babel or you can use async/await, you can get the same result with less code(no catch/then):
// Get reference to all of the documents
console.log("Retrieving list of documents in collection");
let documents = await collectionRef.get();
documents.forEach(async doc => {
console.log("Parent Document ID: ", doc.id);
let subCollectionDocs = await collectionRef.doc(doc.id).collection("subCollection").get()
subCollectionDocs.forEach(subCollectionDoc => {
subCollectionDoc.forEach(doc => {
console.log("Sub Document ID: ", doc.id);
})
});
});
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 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.
I want to provide my users with an API (pointing to my server) that will fetch data from Firebase and return it to them. I want it to be a 'normal' point-in-time request (as opposed to streaming).
My data is 'boxes' within 'projects'. A user can query my API to get all boxes for a project.
My data is normalised, so I will look up the project and get a list of keys of boxes in that project, then go get each box record individually. Once I have them all, I will return the array to the user.
My question: what is the best way to do this?
Here's what I have, and it works. But it feels so hacky.
const projectId = req.params.projectId; // this is passed in by the user in their call to my server.
const boxes = [];
let totalBoxCount = 0;
let fetchedBoxCount = 0;
const projectBoxesRef = db
.child('data/projects')
.child(projectId)
.child('boxes'); // a list of box keys
function getBox(boxSnapshot) {
totalBoxCount++;
db
.child('data/boxes') // a list of box objects
.child(boxSnapshot.key())
.once('value')
.then(boxSnapshot => {
boxes.push(boxSnapshot.val());
fetchedBoxCount++;
if (fetchedBoxCount === totalBoxCount) {
res.json(boxes); // leap of faith that getBox() has been called for all boxes
}
});
}
projectBoxesRef.on('child_added', getBox);
// 'value' fires after all initial 'child_added' things are done
projectBoxesRef.once('value', () => {
projectBoxesRef.off('child_added', getBox);
});
There are some other questions/answers on separating the initial set of child_added objects, and they have influenced my current decision, but they don't seem to relate directly.
Thanks a truck-load for any help.
Update: JavaScript version of Jay's answer below:
db
.child('data/boxes')
.orderByChild(`projects/${projectId}`)
.equalTo(true)
.once('value', boxSnapshot => {
const result = // some parsing of response
res.json(result);
});
This may be too simple a solution but if you have projects, and each project has boxes
your projects node
projects
project_01
boxes
box_id_7: true
box_id_9: true
box_id_34: true
project_37
boxes
box_id_7: true
box_id_14: true
box_id_42: true
and the boxes node
boxes
box_id_7
name: "a 3D box"
shape: "Parallelepiped"
belongs_to_project
project_01: true
box_id_14
name: "I have unequal lenghts"
shape: "Rhumboid"
belongs_to_project
project_37: true
box_id_34
name: "Kinda like a box but with rectangles"
shape: "cuboid"
belongs_to_project
project_01: true
With that, just one (deep) query on the boxes node will load all of the boxes that belong to project_01, which in this case is box_id_7 and box_id_34.
You could go the the other way and since you know the box id for each project in the projects node, you could do a series of observers to load in each project via it's specific path /boxes/box_id_7 etc. I like the query better; faster and less bandwidth.
You could expand on this if a box can belong to multiple projects:
box_id_14
name: "I have unequal lenghts"
shape: "Rhumboid"
belongs_to_project
project_01: true
project_37: true
Now query on the boxes node for all boxes that are part of project_01 will get box_id_7, box_id_14 and box_id_34.
Edit:
Once that structure is in place, use a Deep Query to then get the boxes that belong to the project in question.
For example: suppose you want to craft a Firebase Deep Query to return all boxes where the box's belongs_to_project list contains an item with key "project_37"
boxesRef.queryOrderedByChild("belongs_to_project/project_37"
.queryEqualToValue(true)
.observeSingleEventOfType(.Value, withBlock: { snapshot in
print(snapshot)
})
OK I think I'm happy with my approach, using Promise.all to respond once all the individual 'queries' are returned:
I've changed my approach to use promises, then call Promise.all() to indicate that all the data is ready to send.
const projectId = req.params.projectId;
const boxPromises = [];
const projectBoxesRef = db
.child('data/projects')
.child(projectId)
.child('boxes');
function getBox(boxSnapshot) {
boxPromises.push(db
.child('data/boxes')
.child(boxSnapshot.key())
.once('value')
.then(boxSnapshot => boxSnapshot.val())
);
}
projectBoxesRef.on('child_added', getBox);
projectBoxesRef.once('value', () => {
projectBoxesRef.off('child_added', getBox);
Promise.all(boxPromises).then(boxes => res.json(boxes));
});