I cannot figure out how to proceed with an Upsert & "multiple" onConflict constraints. I want to push a data batch in a Supabase table.
My data array would be structured as follows:
items = [
{ date: "2023-01-26", url: "https://wwww.hello.com"},
{ date: "2023-01-26", url: "https://wwww.goodbye.com"},
...]
I would like to use the Upsert method to push this new batch in my Supabase table, unless if it already exists. To check if it already exists, I would like to use the date, and the url as onConflict criteria, if I understood well.
When I'm running this method
const { error } = await supabase
.from('items')
.upsert(items, { onConflict: ['date','url'] })
.select();
I'm having the following error:
{
code: '42P10',
details: null,
hint: null,
message: 'there is no unique or exclusion constraint matching the ON CONFLICT specification'
}
What am I missing? Where am I wrong?
You can pass more than one column in the upsert into by adding a column in a string (instead of using an array):
const { data, error } = await supabase
.from('items')
.upsert(items, { onConflict: 'date, url'} )
Postgres performs unique index inference as mentioned in https://www.postgresql.org/docs/current/sql-insert.html#SQL-ON-CONFLICT
It is necessary to have unique or indexes for this to work, as you can read in the documentation above:
INSERT into tables that lack unique indexes will not be blocked by
concurrent activity. Tables with unique indexes might block if
concurrent sessions perform actions that lock or modify rows matching
the unique index values being inserted; the details are covered in
Section 64.5. ON CONFLICT can be used to specify an alternative action
to raising a unique constraint or exclusion constraint violation
error.
our current setup is: SPA frontend, Azure functions with mongoose middleware, MongoDB
(Maybe first read the question***)
Since we have a lot of documents in our DB and our customer wants to query them we are facing the following problem:
The user is assigned to his organization. He wants to search for Doc1s he has not responded to.
Doc1
{
_id
organization -> partitionKey
content
}
By creating doc2 with reference to doc1 he can respond.
Doc2
{
_id
organization -> partitionKey
Doc1ref
content
}
We have a 1:n relationship.
At the moment we filter just by query criteria of doc1 with limit and skip options.
But the new requirement is to filter the same way by referring doc2s.
I was thinking of:
Doing it in my code => Problem: after we have read with limit=100 and I filter it by my code, the result is not 100 anymore.
Extending doc1 by doc2 arrays => Must be the last option
Dynamic aggregation, Prepared in the code and executed at runtime => Don't want to user dynamic aggregations and the benefits of mongoose are almost lost.
Create a MongoDB view with lookup aggregation (populating doc1 by doc1.respondedOrganizations) => Problem is see here is the performance. When searching a lot of documents and then joining them by a non partitionKey.
*** So, I come to my question:
Is it possible to pass a virtual (not existing) query criteria...
doc1.find({ alreadyResponded : my.organization } )
...and use it as input variable in an aggregation
{
$lookup: {
from: Doc2s,
localField: _id,
foreignField: Doc1ref,
as: < output array field >
pipeline: [{
$match: {
$organization: {
$eq: $$alreadyResponded
}]
}
}
It would reduce query performance extremly.
Thanks
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.
Im trying to allow the document reading for only users who have in their custom claim the "houseId" that is identical to the document id.
This is the Firestore rule that is not working:
rules_version = '2';
service cloud.firestore {
match /databases/{database}/documents {
match /houses/{houseId} {
allow read: if request.auth.token.houseId == houseId;
}
}
}
I also try the following code, but it didn't work either:
match /houses/{houseId} {
allow read: if request.auth.token.houseId == resource.data.houseId;
}
Instead, if I use the "string" version of the houseId it works.
match /houses/{houseId} {
allow read: if request.auth.token.houseId == 'T45bpx2wzskdtxoowfIw';
}
My client side code for fetching the data is (the inCharge field of the house document is just for querying the specific document where the user has been added):
useEffect(() => {
const subscriber = db.collection('houses')
.where('inCharge.inChargeId', '==', user.userId)
.onSnapshot(querySnapshot => {
const houses = [];
querySnapshot.forEach(documentSnapshot => {
houses.push({
...documentSnapshot.data(),
houseId: documentSnapshot.id
});
});
dispatch(housesActions.setHouses(houses));
setIsLoading(false);
});
return () => subscriber();
}, [])
My user claims are for example like this:
User1: {role: 'inCharge', houseId: T45bpx2wzskdtxoowfIw}
user2: {role: 'viewer', houseId: T45bpx2wzskdtxoowfIw}
...
The documents are like this:
houseId:
name: 'Name of the house'
createdAt: timestamp of creation
inCharge: {
name: 'name of the inCharge person'
inChargeId: 'id of the inCharge person'
}
otherFIelds:
...
Your query does not work because Firestore security rules are not filters. Be sure to read that documentation, and this blog.
It seems that you're depending on the rules to filter out documents that match the user's custom claims. This isn't going to work. Your query needs to apply the filter, and the rules can just check to make sure that the filter is correct.
In your specific case, it looks like each user can access no more than exactly one document in the houses collection, identified by the houseId in their custom claims. Since that's the case, the best thing you can do here is simply get() that document by its ID instead of querying using the other filter that you have now. You can then check that document to see if it contains the values you expect.
You can also try to use the same query you have now, but filter the documentId for the one you provide, which must match the one in custom claims:
.where(firebase.FieldValue.documentId(), '==', houseId)
But this seems more complicated to me since your rules require that each user can only read their one document.
I want to obtain all the fields of a schema in mongoose. Now I am using the following code:
let Client = LisaClient.model('Client', ClientSchema)
let query = Client.findOne({ 'userclient': userclient })
query.select('clientname clientdocument client_id password userclient')
let result = yield query.exec()
But I want all the fields no matter if they are empty. As always, in advance thank you
I'm not sure if you want all fields in a SQL-like way, or if you want them all in a proper MongoDB way.
If you want them in the proper MongoDB way, then just remove the query.select line. That line is saying to only return the fields listed in it.
If you meant in a SQL-like way, MongoDB doesn't work like that. Each document only has the fields you put in when it was inserted. If when you inserted the document, you only gave it certain fields, that document will only have those fields, even if other documents in other collections have different fields.
To determine all available fields in the collection, you'd have to find all the documents, loop through them all and build an object with all the different keys you find.
If you need each document returned to always have the fields that you specify in your select, you'll just have to transform your object once it's returned.
const fields = ['clientname', 'clientdocument', 'client_id', 'password', 'userclient'];
let Client = LisaClient.model('Client', ClientSchema)
let query = Client.findOne({ 'userclient': userclient })
query.select(fields.join(' '))
let result = yield query.exec()
fields.forEach(field => result[field] = result[field]);
That forEach loop will set all the fields you want to either the value in the result (if it was there) or to undefined if it wasn't.
MongoDB is schemaless and does not have tables, each collection can have different types of items.Usually the objects are somehow related or have a common base type.
Retrive invidual records using
db.collectionName.findOne() or db.collectionName.find().pretty()
To get all key names you need to MapReduce
mapReduceKeys = db.runCommand({
"mapreduce": "collection_name",
"map": function() {
for (var key in this) {
emit(key, null);
}
},
"reduce": function(key, stuff) {
return null;
},
"out": "collection_name" + "_keys"
})
Then run distinct on the resulting collection so as to find all the keys
db[mapReduceKeys.result].distinct("_id") //["foo", "bar", "baz", "_id", ...]