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Web Applications

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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;

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