This query is marked as a draft This query has been published by TanvirSdq.

SQL

x
 
SELECT
    a.actor_user AS user_id,
    u.user_name,
    COUNT(*) AS talk_edits
FROM revision r
JOIN actor a ON r.rev_actor = a.actor_id
JOIN user u ON a.actor_user = u.user_id
JOIN page p ON r.rev_page = p.page_id
WHERE p.page_namespace % 2 != 0  -- All talk pages (namespace is odd)
GROUP BY a.actor_user, u.user_name
HAVING COUNT(*) >= 100
ORDER BY talk_edits DESC;
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