mysql - SQL选择前百分之n行

我有一个相当困难的问题。
我需要在一个名为'key_phrases'（columns:'user'，'phrase'，'tfw'）的表上编写一个查询，该表检索每个用户前10%的行，这些行的最大值为'tfw'列。查询应返回所有列。
提前谢谢你，尼尔

最佳答案

select user, phrase, tfw
from key_phrases t1
join (
    select count(*) total_rows_per_user, user
    from key_phrases
    group by user
) t2 on t1.user = t2.user
where (
    select count(*) from key_phrases t3
    where t3.user = t1.user
    and t3.tfw >= t1.tfw
) / total_rows_per_user <= .1

另一个使用变量的查询应该更快

select user, phrase, tfw,
if(@prev_user = user, @user_count := @user_count + 1, @user_count := 1),
@prev_user := user
from key_phrases t1
join (
    select count(*) total_rows_per_user, user
    from key_phrases
    group by user
) t2 on t1.user = t2.user
cross join (select @user_count := 1, @prev_user := null) t3
where @user_count / total_rows_per_user >= .9
order by user, tfw

关于mysql - SQL选择前百分之n行，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/25549172/