Related
From the docs:
You can also chain multiple where() methods to create more specific queries (logical AND).
How can I perform an OR query?
Example:
Give me all documents where the field status is open OR upcoming
Give me all documents where the field status == open OR createdAt <= <somedatetime>
OR isn't supported as it's hard for the server to scale it (requires keeping state to dedup). The work around is to issue 2 queries, one for each condition, and dedup on the client.
Edit (Nov 2019):
Cloud Firestore now supports IN queries which are a limited type of OR query.
For the example above you could do:
// Get all documents in 'foo' where status is open or upcmoming
db.collection('foo').where('status','in',['open','upcoming']).get()
However it's still not possible to do a general OR condition involving multiple fields.
With the recent addition of IN queries, Firestore supports "up to 10 equality clauses on the same field with a logical OR"
A possible solution to (1) would be:
documents.where('status', 'in', ['open', 'upcoming']);
See Firebase Guides: Query Operators | in and array-contains-any
suggest to give value for status as well.
ex.
{ name: "a", statusValue = 10, status = 'open' }
{ name: "b", statusValue = 20, status = 'upcoming'}
{ name: "c", statusValue = 30, status = 'close'}
you can query by ref.where('statusValue', '<=', 20) then both 'a' and 'b' will found.
this can save your query cost and performance.
btw, it is not fix all case.
I would have no "status" field, but status related fields, updating them to true or false based on request, like
{ name: "a", status_open: true, status_upcoming: false, status_closed: false}
However, check Firebase Cloud Functions. You could have a function listening status changes, updating status related properties like
{ name: "a", status: "open", status_open: true, status_upcoming: false, status_closed: false}
one or the other, your query could be just
...where('status_open','==',true)...
Hope it helps.
This doesn't solve all cases, but for "enum" fields, you can emulate an "OR" query by making a separate boolean field for each enum-value, then adding a where("enum_<value>", "==", false) for every value that isn't part of the "OR" clause you want.
For example, consider your first desired query:
Give me all documents where the field status is open OR upcoming
You can accomplish this by splitting the status: string field into multiple boolean fields, one for each enum-value:
status_open: bool
status_upcoming: bool
status_suspended: bool
status_closed: bool
To perform your "where status is open or upcoming" query, you then do this:
where("status_suspended", "==", false).where("status_closed", "==", false)
How does this work? Well, because it's an enum, you know one of the values must have true assigned. So if you can determine that all of the other values don't match for a given entry, then by deduction it must match one of the values you originally were looking for.
See also
in/not-in/array-contains-in: https://firebase.google.com/docs/firestore/query-data/queries#in_and_array-contains-any
!=: https://firebase.googleblog.com/2020/09/cloud-firestore-not-equal-queries.html
I don't like everyone saying it's not possible.
it is if you create another "hacky" field in the model to build a composite...
for instance, create an array for each document that has all logical or elements
then query for .where("field", arrayContains: [...]
you can bind two Observables using the rxjs merge operator.
Here you have an example.
import { Observable } from 'rxjs/Observable';
import 'rxjs/add/observable/merge';
...
getCombinatedStatus(): Observable<any> {
return Observable.merge(this.db.collection('foo', ref => ref.where('status','==','open')).valueChanges(),
this.db.collection('foo', ref => ref.where('status','==','upcoming')).valueChanges());
}
Then you can subscribe to the new Observable updates using the above method:
getCombinatedStatus.subscribe(results => console.log(results);
I hope this can help you, greetings from Chile!!
We have the same problem just now, luckily the only possible values for ours are A,B,C,D (4) so we have to query for things like A||B, A||C, A||B||C, D, etc
As of like a few months ago firebase supports a new query array-contains so what we do is make an array and we pre-process the OR values to the array
if (a) {
array addObject:#"a"
}
if (b) {
array addObject:#"b"
}
if (a||b) {
array addObject:#"a||b"
}
etc
And we do this for all 4! values or however many combos there are.
THEN we can simply check the query [document arrayContains:#"a||c"] or whatever type of condition we need.
So if something only qualified for conditional A of our 4 conditionals (A,B,C,D) then its array would contain the following literal strings: #["A", "A||B", "A||C", "A||D", "A||B||C", "A||B||D", "A||C||D", "A||B||C||D"]
Then for any of those OR combinations we can just search array-contains on whatever we may want (e.g. "A||C")
Note: This is only a reasonable approach if you have a few number of possible values to compare OR with.
More info on Array-contains here, since it's newish to firebase docs
If you have a limited number of fields, definitely create new fields with true and false like in the example above. However, if you don't know what the fields are until runtime, you have to just combine queries.
Here is a tags OR example...
// the ids of students in class
const students = [studentID1, studentID2,...];
// get all docs where student.studentID1 = true
const results = this.afs.collection('classes',
ref => ref.where(`students.${students[0]}`, '==', true)
).valueChanges({ idField: 'id' }).pipe(
switchMap((r: any) => {
// get all docs where student.studentID2...studentIDX = true
const docs = students.slice(1).map(
(student: any) => this.afs.collection('classes',
ref => ref.where(`students.${student}`, '==', true)
).valueChanges({ idField: 'id' })
);
return combineLatest(docs).pipe(
// combine results by reducing array
map((a: any[]) => {
const g: [] = a.reduce(
(acc: any[], cur: any) => acc.concat(cur)
).concat(r);
// filter out duplicates by 'id' field
return g.filter(
(b: any, n: number, a: any[]) => a.findIndex(
(v: any) => v.id === b.id) === n
);
}),
);
})
);
Unfortunately there is no other way to combine more than 10 items (use array-contains-any if < 10 items).
There is also no other way to avoid duplicate reads, as you don't know the ID fields that will be matched by the search. Luckily, Firebase has good caching.
For those of you that like promises...
const p = await results.pipe(take(1)).toPromise();
For more info on this, see this article I wrote.
J
OR isn't supported
But if you need that you can do It in your code
Ex : if i want query products where (Size Equal Xl OR XXL : AND Gender is Male)
productsCollectionRef
//1* first get query where can firestore handle it
.whereEqualTo("gender", "Male")
.addSnapshotListener((queryDocumentSnapshots, e) -> {
if (queryDocumentSnapshots == null)
return;
List<Product> productList = new ArrayList<>();
for (DocumentSnapshot snapshot : queryDocumentSnapshots.getDocuments()) {
Product product = snapshot.toObject(Product.class);
//2* then check your query OR Condition because firestore just support AND Condition
if (product.getSize().equals("XL") || product.getSize().equals("XXL"))
productList.add(product);
}
liveData.setValue(productList);
});
For Flutter dart language use this:
db.collection("projects").where("status", whereIn: ["public", "unlisted", "secret"]);
actually I found #Dan McGrath answer working here is a rewriting of his answer:
private void query() {
FirebaseFirestore db = FirebaseFirestore.getInstance();
db.collection("STATUS")
.whereIn("status", Arrays.asList("open", "upcoming")) // you can add up to 10 different values like : Arrays.asList("open", "upcoming", "Pending", "In Progress", ...)
.addSnapshotListener(new EventListener<QuerySnapshot>() {
#Override
public void onEvent(#Nullable QuerySnapshot queryDocumentSnapshots, #Nullable FirebaseFirestoreException e) {
for (DocumentSnapshot documentSnapshot : queryDocumentSnapshots) {
// I assume you have a model class called MyStatus
MyStatus status= documentSnapshot.toObject(MyStatus.class);
if (status!= null) {
//do somthing...!
}
}
}
});
}
I'd like to determine the number of assigned companies to a category. The (simplified) data structure of the input topic looks something like this:
Key: 3298440
Value: {"company_id": 5678, "category_id": 9876}
Key: 4367848
Value: {"company_id": 35383, "category_id": 9876}
[...]
So, like in this example, I'd like to count the different companies for the category 9876.
My idea was to group by category_id (the new key) and in- or decrease the count (via reduce()) as the value.
So in principle, I'm only interested in inserts and deletes (tombstones) for altering the count. My problem is, that updates will also produce a record and invalidate the count.
I believe I'm on the wrong track here. Is there any way to do it?
This is my incorrect java code, which also counts updates:
companyCategoriesKTable
.groupBy(KeyValue::pair, Grouped.with(Serdes.String(), companyCategoriesSerde))
.aggregate(
() -> new CompanyCount(0L, 0L), /* category_id, count */
(key, newValue, aggValue) -> {
aggValue.setCategoryId(newValue.getBrId());
aggValue.setCount(aggValue.getCount() + 1);
return aggValue;
},
(key, oldValue, aggValue) -> {
aggValue.setCategoryId(oldValue.getBrId());
aggValue.setCount(aggValue.getCount() - 1);
return aggValue;
},
Materialized.<String, CompanyCount, KeyValueStore<Bytes, byte[]>>as("CompanyCountStore").withKeySerde(Serdes.String()).withValueSerde(new JsonSerde<>(CompanyCount.class))
)
.toStream()
.map((k, v) -> new KeyValue<>(v.getCategoryId(), v.getCount() > 0 ? 1L : -1L))
.groupByKey(Grouped.with(Serdes.Long(), Serdes.Long()))
.reduce(Long::sum)
.toStream()
.to("output-topic", Produced.with(Serdes.Long(), Serdes.Long()));
I would like to do the following with RxJava
class Invoice(val dayOfMonth:Int,val amount:Int)
below is the sample monthInvoices:List< Invoice > to process
Invoice(3,100)
Invoice(3,150)
Invoice(3,50)
Invoice(4,350)
Invoice(8,400)
Invoice(8,100)
First, I would like to group it by the day of the month like the following
Invoice(3,300)
Invoice(4,350)
Invoice(8,500)
Then I would like to create a list containing all the days of the month. Say, we are having 30 days for this month, then the output list must contain inserting a empty Invoice object with 0 amount for the days where there is no invoice
Desired output List
Invoice(1,0) //Since day 1 is not in the group summed list
Invoice(2,0) //day 2 is also not there
Invoice(3,300)
Invoice(4,350)
Invoice(5,0)
Invoice(6,0)
Invoice(7,0)
Invoice(8,500)
…..
Invoice(30,0)
Hope I have explained the need clearly. Can anyone please answer me a solution to do it entirely using RxJava?
Try this
fun task(invoices: List<Invoice>) =
Observable.fromIterable(invoices)
.groupBy { it.dayOfMonth }
.flatMapSingle { group -> group.reduce(0) { t1, t2 -> t1 + t2.amount }
.map { group.key to it }}
.toMap({ it.first }, { it.second })
.flatMapObservable { map ->
Observable.range(1, 30)
.map { Invoice(it, map[it] ?: 0) }
}
This can be achieved much more easily using the collection operators inside Kotlin's standard library, but in pure RxJava you can do this by using groupBy and reduce.
val invoices = listOf(
Invoice(3, 100),
Invoice(3, 150),
Invoice(3, 50),
Invoice(4, 350),
Invoice(8, 400),
Invoice(8, 100)
)
Observable.range(1, 30)
.map { Invoice(it, 0) } // Create an Observable of Invoice([day], 0)
.mergeWith(Observable.fromIterable(invoices))
.groupBy { it.dayOfMonth } // Merge the sources and groupBy day
.flatMapMaybe { group ->
group.reduce { t1: Invoice, t2: Invoice ->
Invoice(t1.dayOfMonth, t1.amount + t2.amount) // Reduce each group into a single Invoice
}
}
.subscribe {
// Optionally you can call toList before this if you want to aggregate the emissions into a single list
println(it)
}
From the docs:
You can also chain multiple where() methods to create more specific queries (logical AND).
How can I perform an OR query?
Example:
Give me all documents where the field status is open OR upcoming
Give me all documents where the field status == open OR createdAt <= <somedatetime>
OR isn't supported as it's hard for the server to scale it (requires keeping state to dedup). The work around is to issue 2 queries, one for each condition, and dedup on the client.
Edit (Nov 2019):
Cloud Firestore now supports IN queries which are a limited type of OR query.
For the example above you could do:
// Get all documents in 'foo' where status is open or upcmoming
db.collection('foo').where('status','in',['open','upcoming']).get()
However it's still not possible to do a general OR condition involving multiple fields.
With the recent addition of IN queries, Firestore supports "up to 10 equality clauses on the same field with a logical OR"
A possible solution to (1) would be:
documents.where('status', 'in', ['open', 'upcoming']);
See Firebase Guides: Query Operators | in and array-contains-any
suggest to give value for status as well.
ex.
{ name: "a", statusValue = 10, status = 'open' }
{ name: "b", statusValue = 20, status = 'upcoming'}
{ name: "c", statusValue = 30, status = 'close'}
you can query by ref.where('statusValue', '<=', 20) then both 'a' and 'b' will found.
this can save your query cost and performance.
btw, it is not fix all case.
I would have no "status" field, but status related fields, updating them to true or false based on request, like
{ name: "a", status_open: true, status_upcoming: false, status_closed: false}
However, check Firebase Cloud Functions. You could have a function listening status changes, updating status related properties like
{ name: "a", status: "open", status_open: true, status_upcoming: false, status_closed: false}
one or the other, your query could be just
...where('status_open','==',true)...
Hope it helps.
This doesn't solve all cases, but for "enum" fields, you can emulate an "OR" query by making a separate boolean field for each enum-value, then adding a where("enum_<value>", "==", false) for every value that isn't part of the "OR" clause you want.
For example, consider your first desired query:
Give me all documents where the field status is open OR upcoming
You can accomplish this by splitting the status: string field into multiple boolean fields, one for each enum-value:
status_open: bool
status_upcoming: bool
status_suspended: bool
status_closed: bool
To perform your "where status is open or upcoming" query, you then do this:
where("status_suspended", "==", false).where("status_closed", "==", false)
How does this work? Well, because it's an enum, you know one of the values must have true assigned. So if you can determine that all of the other values don't match for a given entry, then by deduction it must match one of the values you originally were looking for.
See also
in/not-in/array-contains-in: https://firebase.google.com/docs/firestore/query-data/queries#in_and_array-contains-any
!=: https://firebase.googleblog.com/2020/09/cloud-firestore-not-equal-queries.html
I don't like everyone saying it's not possible.
it is if you create another "hacky" field in the model to build a composite...
for instance, create an array for each document that has all logical or elements
then query for .where("field", arrayContains: [...]
you can bind two Observables using the rxjs merge operator.
Here you have an example.
import { Observable } from 'rxjs/Observable';
import 'rxjs/add/observable/merge';
...
getCombinatedStatus(): Observable<any> {
return Observable.merge(this.db.collection('foo', ref => ref.where('status','==','open')).valueChanges(),
this.db.collection('foo', ref => ref.where('status','==','upcoming')).valueChanges());
}
Then you can subscribe to the new Observable updates using the above method:
getCombinatedStatus.subscribe(results => console.log(results);
I hope this can help you, greetings from Chile!!
We have the same problem just now, luckily the only possible values for ours are A,B,C,D (4) so we have to query for things like A||B, A||C, A||B||C, D, etc
As of like a few months ago firebase supports a new query array-contains so what we do is make an array and we pre-process the OR values to the array
if (a) {
array addObject:#"a"
}
if (b) {
array addObject:#"b"
}
if (a||b) {
array addObject:#"a||b"
}
etc
And we do this for all 4! values or however many combos there are.
THEN we can simply check the query [document arrayContains:#"a||c"] or whatever type of condition we need.
So if something only qualified for conditional A of our 4 conditionals (A,B,C,D) then its array would contain the following literal strings: #["A", "A||B", "A||C", "A||D", "A||B||C", "A||B||D", "A||C||D", "A||B||C||D"]
Then for any of those OR combinations we can just search array-contains on whatever we may want (e.g. "A||C")
Note: This is only a reasonable approach if you have a few number of possible values to compare OR with.
More info on Array-contains here, since it's newish to firebase docs
If you have a limited number of fields, definitely create new fields with true and false like in the example above. However, if you don't know what the fields are until runtime, you have to just combine queries.
Here is a tags OR example...
// the ids of students in class
const students = [studentID1, studentID2,...];
// get all docs where student.studentID1 = true
const results = this.afs.collection('classes',
ref => ref.where(`students.${students[0]}`, '==', true)
).valueChanges({ idField: 'id' }).pipe(
switchMap((r: any) => {
// get all docs where student.studentID2...studentIDX = true
const docs = students.slice(1).map(
(student: any) => this.afs.collection('classes',
ref => ref.where(`students.${student}`, '==', true)
).valueChanges({ idField: 'id' })
);
return combineLatest(docs).pipe(
// combine results by reducing array
map((a: any[]) => {
const g: [] = a.reduce(
(acc: any[], cur: any) => acc.concat(cur)
).concat(r);
// filter out duplicates by 'id' field
return g.filter(
(b: any, n: number, a: any[]) => a.findIndex(
(v: any) => v.id === b.id) === n
);
}),
);
})
);
Unfortunately there is no other way to combine more than 10 items (use array-contains-any if < 10 items).
There is also no other way to avoid duplicate reads, as you don't know the ID fields that will be matched by the search. Luckily, Firebase has good caching.
For those of you that like promises...
const p = await results.pipe(take(1)).toPromise();
For more info on this, see this article I wrote.
J
OR isn't supported
But if you need that you can do It in your code
Ex : if i want query products where (Size Equal Xl OR XXL : AND Gender is Male)
productsCollectionRef
//1* first get query where can firestore handle it
.whereEqualTo("gender", "Male")
.addSnapshotListener((queryDocumentSnapshots, e) -> {
if (queryDocumentSnapshots == null)
return;
List<Product> productList = new ArrayList<>();
for (DocumentSnapshot snapshot : queryDocumentSnapshots.getDocuments()) {
Product product = snapshot.toObject(Product.class);
//2* then check your query OR Condition because firestore just support AND Condition
if (product.getSize().equals("XL") || product.getSize().equals("XXL"))
productList.add(product);
}
liveData.setValue(productList);
});
For Flutter dart language use this:
db.collection("projects").where("status", whereIn: ["public", "unlisted", "secret"]);
actually I found #Dan McGrath answer working here is a rewriting of his answer:
private void query() {
FirebaseFirestore db = FirebaseFirestore.getInstance();
db.collection("STATUS")
.whereIn("status", Arrays.asList("open", "upcoming")) // you can add up to 10 different values like : Arrays.asList("open", "upcoming", "Pending", "In Progress", ...)
.addSnapshotListener(new EventListener<QuerySnapshot>() {
#Override
public void onEvent(#Nullable QuerySnapshot queryDocumentSnapshots, #Nullable FirebaseFirestoreException e) {
for (DocumentSnapshot documentSnapshot : queryDocumentSnapshots) {
// I assume you have a model class called MyStatus
MyStatus status= documentSnapshot.toObject(MyStatus.class);
if (status!= null) {
//do somthing...!
}
}
}
});
}
Scan works like this(with sum function):
1-1-1-1-1-1 -> 1-2-3-4-5-6
but I need something like this: with n=3
1-1-1-1-1-1 -> 3-3
How do I achieve this behavior?
In RxJs you can use bufferWithCount:
var source = Rx.Observable.from([1,1,1,1,1,1])
.bufferWithCount(3)
.flatMap(group => Rx.Observable.from(group).sum());
You could optionally use windowWithCount so that you don't have to rewrap the output in order to use sum but you'll also get an empty final window as well, which will fire out an extraneous 0 value.
This works in c#:
var query =
new [] { 1, 1, 1, 1, 1, 1, }
.ToObservable()
.Buffer(3)
.Select(x => x.Sum());
I assume [rxjs] would work similarly.