Re: SQL question, TOP 5 and all OTHERS
От
Skylar Thompson
Тема
Re: SQL question, TOP 5 and all OTHERS
Дата
Msg-id
20220606211213.dntvfnkwauk52fzc@thargelion
Ответ на
Re: SQL question, TOP 5 and all OTHERS (Jean MAURICE)
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Re: SQL question, TOP 5 and all OTHERS Skylar Thompson <skylar2@uw.edu>
On Mon, Jun 06, 2022 at 09:46:12PM +0200, Jean MAURICE wrote: > Hi Scott, > what about using a Common Table Expression and the clause WITH ? > I am not at home now but you can write something like > > WITH top5 AS (SELECT vendor_name AS vendor_name, > > ?????? count(DISTINCT inv_id) AS "# of Invoices" > > FROM SpendTable > > GROUP BY vendor_name > > ORDER BY "# of Invoices" DESC > > LIMIT 5) > SELECT * FROM top5 > UNION > > SELECT 'all other' AS vendor_name, > > ?????? count(DISTINCT st.inv_id) AS "# of Invoices" > > FROM SpendTable AS st > > WHERE st.vendor_name NOT IN (SELECT vendor_name FROM top5) > > ORDER BY "# of Invoices" DESC There might be a challenge with ties, especially if you don't order by the vendor name since you could get different results even on the same data set, depending on how the query plan goes. It depends on what the OP is looking for, I guess. -- -- Skylar Thompson (skylar2@u.washington.edu) -- Genome Sciences Department (UW Medicine), System Administrator -- Foege Building S046, (206)-685-7354 -- Pronouns: He/Him/His
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