select t2.date, t2.user_id, t2.sum_pass_count as pass_count
from (select t1.date, t1.user_id, t1.sum_pass_count,
    dense_rank() over (partition by t1.date order by t1.sum_pass_count desc) rn-- 不跳数字排名
    from 
        (select date, user_id, sum(pass_count) as sum_pass_count
        from questions_pass_record
        group by user_id, date) as t1-- 考虑一个用户在同一天内刷了不同题型,所以结果前二的user-id不会是同一个人
    ) as t2
where t2.rn <= 2
order by t2.date