$query1 = select disinct(b2.username)
from bookmark b1, bookmark b2
where b1.usernam = '$valid_user'
and b1.username != b2.username
and b1.bm_url = b2.bm_url
該查詢將給出一個與當前用戶意向相似的人的列表
$query2 = select bm_url
from bookmark
where username in
($query1)
假如用戶已經有了該書籤，就不要再推薦給他
and bm_url not in
(select bm_url from bookmark where username="$valid_user")
並且只推薦大衆化的的標籤，不推薦太個性化的標籤
group by bm_url
having count(bm_url) > $popularity
