How to use multiple where clause in Flutter Firestore [duplicate]

Question

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:

1. Give me all documents where the field status is open OR upcoming
2. Give me all documents where the field status == open OR createdAt <= <somedatetime>

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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!!

Answer 6

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

Answer 7

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:

1. 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

Answer 8

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

Answer 9

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: [...]