题解 | #每篇文章同一时刻最大在看人数#

### 不好意思又是一篇照搬别人代码的行为。一开始我的思路出劈叉了，我想着通过先计算某个瞬时时刻，然后通过瞬时时刻分组计算每个时刻的进出总人数，再比较出每篇文章的最大进出总人数；参考了别人的写法之后，我顿悟了，将in_time 时间记作1，代表人数增加，将out_time记作-1，代表人数的减少，先统计所有时刻及对应的人数增加及减少，然后根据sum() over() 窗口函数统计每个截止时刻的进出总人数：sum(diff) over(partition by artical_id order by dt,diff desc)


### 每篇文章同一时刻最大在看人数   
 
       # 统计截止到每个时刻 每篇文章的在浏览人数   
       
       select  artical_id,
               max(sumUv) max_uv 
               from (
       select  artical_id,
               dt ,
               sum(diff) over(partition by artical_id order by dt,diff desc) sumUv  # 每篇文章截止时间的进出总人数
         from (
        select artical_id ,
               in_time dt,
               1 diff 
          from tb_user_log where artical_id != 0 
          union all 
        select artical_id,
               out_time dt,
               -1 diff 
          from tb_user_log where artical_id != 0 
          ) A 
          ) B group by B.artical_id order by max_uv desc ;