How to perform compound queries with logical OR in Cloud Firestore?

Question

From the docs:

You can also chain multiple where() methods to create more specific queries (logical AND).

How can I perform an <

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

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 3

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.