100高分求sql,该怎么处理

100高分求sql

数据库 mysql

表pic(pic_is,user_id,comment_num,praise_num,browser_num,hot_status)

hot_status 为图片状态（1，2，3），其中3表示是热图。

我想查出用户id（去重）列表，并且是按这个计算权重的倒序排序，

计算公式：comment_num*2+praise_num+browser_num*0.2+热图数*3

就是说按某个用户的所有图片的评论总数*2加上他的所有图片的赞数，加上他的所有图片的浏览数*0.2，再加上他的所有热图数*3，按这个计算出来的值倒序排序……

不知道我说的明白不。

在线，等高手回答，，，，，

跪求……

SQL code

select *
from (
select userid,sum(comment_num)*2+sum(praise_num)+sum(browser_num)*0.2+sum(热图数)*3 as quanzhong
from tb
group by user_id
) D
order by quanzhong desc
limit 10;



  
  

   
   SQL code
   
   
select user_id, sum(comment_num*2+praise_num+browser_num*0.2+热图数*3)
form pic
group by user_id
order by 2 desc



    
     探讨
     

     
SQL code
     
select *
     
from (
     
select userid,sum(comment_num)*2+sum(praise_num)+sum(browser_num)*0.2+sum(热图数)*3 as quanzhong
     
from tb
     
group by user_id
     
) D
     
order by quanzhong desc
     
limit 10;
     

    


贴建表及插入记录的SQL，及要求结果出来看看


select userid,sum(comment_num)*2+sum(praise_num)+sum(browser_num)*0.2+sum(热图数)*3 as quanzhong
from tb
group by user_id order by 2 desc



    
    

     
     SQL code
     
     
SELECT  userid,SUM(comment_num)*2+SUM(praise_num)+SUM(browser_num)*0.2+SUM(热图数)*3 as quanzhong
FROM  tb
GROUP BY  user_id ORDER BY  2 DESC;


热图数要另外统计的吧。。


select userid,sum(comment_num)*2+sum(praise_num)+sum(browser_num)*0.2+
sum(if(hot_status=3,1,0))*3 as quanzhong
from tb
group by user_id order by 2 desc



      
       探讨
       

       
select userid,sum(comment_num)*2+sum(praise_num)+sum(browser_num)*0.2+
       
sum(if(hot_status=3,1,0))*3 as quanzhong
       
from tb
       
group by user_id order by 2 desc