Programming

I'm too lazy to convert that by hand, but here's what chatgpt converted that to for SQL, for the sake of discussion:

SELECT 
    a.id,
    a.artist_name -- or whatever the name column is in the 'artists' table
FROM artists a
JOIN albums al ON a.id = al.artist_id
JOIN nominations n ON al.id = n.album_id -- assuming nominations are for albums
WHERE al.release_date BETWEEN '1990-01-01' AND '1999-12-31'
AND n.award = 'MTV' -- assuming there's a column that specifies the award name
AND n.won = FALSE
GROUP BY a.id, a.artist_name -- or whatever the name column is in the 'artists' table
ORDER BY COUNT(DISTINCT n.id) DESC, a.artist_name -- ordering by the number of nominations, then by artist name
LIMIT 10;

I like Django's ORM just fine, but that SQL isn't too bad (it's also slightly different than your version though, but works fine as an example). I also like PyPika sometimes for building queries when I'm not using Django or SQLAlchemy, and here's that version:

q = (
    Query
    .from_(artists)
    .join(albums).on(artists.id == albums.artist_id)
    .join(nominations).on(albums.id == nominations.album_id)
    .select(artists.id, artists.artist_name)  # assuming the column is named artist_name
    .where(albums.release_date.between('1990-01-01', '1999-12-31'))
    .where(nominations.award == 'MTV')
    .where(nominations.won == False)
    .groupby(artists.id, artists.artist_name)
    .orderby(fn.Count(nominations.id).desc(), artists.artist_name)
    .limit(10)
)

I think PyPika answers your concerns about

What if one method wants the result of that but only wants the artists’ names, but another one wanted additional or other fields?

It's just regular Python code, same as the Django ORM.