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Non-Blocked "88" users
by SQL
This query is marked as a draft
This query has been published
by Cryptic.
SQL
AخA
SELECT user_name,
user_editcount,
user_registration
FROM user
WHERE (user_name LIKE "%88"
OR user_name LIKE "88%")
AND user_id NOT IN (SELECT ipb_user
FROM ipblocks
WHERE ipb_user = user_id)
ORDER BY user_editcount DESC;
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