Reputation: 60179
I have a legacy MySQL table called lnk_lists_addresses
with columns list_id
and address_id
. I'd like to write a query that reports all the cases where the same list_id
-address_id
combination appears more than once in the table with a count.
I tried this...
SELECT count(*), list_id, address_id
FROM lnk_lists_addresses
GROUP BY list_id, address_id
ORDER BY count(*) DESC
LIMIT 20
It works, sort of, because there are fewer than 20 duplicates. But how would I return only the counts greater than 1?
I tried adding "WHERE count(*) > 1
" before and after GROUP BY
but got errors saying the statement was invalid.
Upvotes: 1
Views: 5200
Reputation: 30006
SELECT count(*), list_id, address_id
FROM lnk_lists_addresses
GROUP BY list_id, address_id
HAVING count(*)>1
ORDER BY count(*) DESC
To combine mine and Todd.Run's answers for a more "complete" answer. You want to use the HAVING clause:
http://dev.mysql.com/doc/refman/5.1/en/select.html
Upvotes: 8
Reputation: 24472
SELECT count(*) AS total, list_id, address_id
FROM lnk_lists_addresses
WHERE total > 1
GROUP BY list_id, address_id
ORDER BY total DESC
LIMIT 20
If you name the COUNT() field, you can use it later in the statement.
EDIT: forgot about HAVING
(>_<)
Upvotes: 1
Reputation: 18526
You want to use a "HAVING" clause. Its use is explained in the MySQL manual.
http://dev.mysql.com/doc/refman/5.1/en/select.html
Upvotes: 2