Re: Most Occurring Value
От | Osvaldo Rosario Kussama |
---|---|
Тема | Re: Most Occurring Value |
Дата | |
Msg-id | 47FB9944.4070307@gmail.com обсуждение исходный текст |
Ответ на | Re: Most Occurring Value (Volkan YAZICI <yazicivo@ttmail.com>) |
Список | pgsql-general |
Volkan YAZICI escreveu: > Mike Ginsburg <mginsburg@collaborativefusion.com> writes: >> There is probably a really simple solution for this problem, but for >> the life of me I can't see to think of it. I have three tables >> >> --contains u/p for all users in the site >> TABLE users (user_id INT primary key, username VARCHAR(50), password TEXT) >> --list of all possible events (login, logout, timeout) >> TABLE events (event_id INT primary key, event VARCHAR(255)) >> --logs the activity of all users logging in/out, etc >> TABLE log (log_id INT primary key, user_id INT REFERENCES users, >> event_id INT REFERENCES event); >> >> How would I query to find out which user has the most activity? >> SELECT user_id, COUNT(event_id) >> FROM log >> GROUP BY (user_id) >> HAVNG COUNT(event_id) = ??? > > SELECT user_id, max(count(event_id)) max(count() is invalid. aggregate function calls may not be nested > FROM log > GROUP BY user_id; > > or > > SELECT user_id, count(event_id) > FROM log > GROUP BY user_id > ORDER BY count(event_id) DESC > LIMIT 1; If more than 1 user has the most activity only one is listed. Try: SELECT user_id, COUNT(event_id) FROM log GROUP BY (user_id) HAVING COUNT(event_id) = (SELECT max(l.ct) FROM (SELECT count(event_id) AS ct FROM log GROUP BY user_id) AS l) ORDER BY user_id; Osvaldo
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