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

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.

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.

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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