Supabase sort / filter by relation count

Question

I have the following query:

let query = supabase
      .from("property_owners")
      .select(
        `id, full_name,  property(id), address(id, street, house_number, city, zipcode, property(id))`,
      );

Now, I want to find a way to either:

Sort the property owners by property count
Filter the property owners by property count (get all property owners with > 5 properties)

I can't seem to find a way to do this, is this a limit of supabase?

You can always use views or rpcs to perform complicated queries instead of directly using the Supabase client library. — dshukertjr
– dshukertjr, Commented May 14, 2024 at 5:32

Arash Torkaman · Accepted Answer · 2025-10-23 07:51:39Z

0

you can user order to sort the returns list

let query = supabase
      .from("property_owners")
      .select()
      .order(columnName, ascending: true/false);

and for find rows that a perperty (numeric) is greater or lessser that a value use :
gt : greater than
gte : greater than or equal
lt : less than
lte : less that or equal

let query = supabase
      .from("property_owners")
      .select()
      .order(columnName, ascending: true/false)
      .gt(count, 5);

answered Oct 23 at 7:51

Arash Torkaman

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