Q&A for power users of web applications
with UserPostCounts as ( select users.displayname, count(posts.owneruserid) as post_count from users, posts where posts.owneruserid = users.id and location = 'Russia' group by users.displayname having count(posts.owneruserid) > some( select count(posts.owneruserid) from users, posts where posts.owneruserid = users.id and location = 'Poland' group by users.displayname)) select top 5 displayname, post_count from UserPostCounts order by post_count desc;