Question

我有一张这样的表

userEmail | views

  A          8
  B          3
  C          4
  A          2
  B          5

我希望获得特定用户的排名说B ..

我正在做的是：

SET @rank=0;
select @rank:=@rank+1 AS rank , userEmail,sum(views) as views from mtable where userEmail != 'null' Group By userEmail Order By views DESC

这将输出如下内容：

rank  | userEmail | views

  1        A          10
  2        B           8
  3        C           4

我可以获得所有用户的排名，但如果我想看到只有一个用户的排名呢？

如果我将where userEmail = 'B'放在上面的查询

中

我会得到这个：

rank  | userEmail | views

  1        B           8

但它应该是这样的：

 rank  | userEmail | views

  2          B           8

Answer 1

正如Gordan指出我的第一个答案不是很标准。这是做同样事情的标准方法。

SELECT T.userEmail, T.views, RANK() OVER (ORDER BY T.views DESC) AS viewRank
FROM (
  select userEmail,sum(views) as views 
  from mtable 
  where userEmail != 'null' 
  Group By userEmail) AS T
Order By T.views DESC

在sql server 2005+和Oracle（使用OVER子句）中有一种特殊的方法可以做到这一点，但是一般的答案（必须与mysql一起使用）是使用where子句。像这样：

SET @rank=0;

SELECT T.rank, T.userEmail, T.views 
FROM (
  select @rank:=@rank+1 AS rank , userEmail,sum(views) as views 
  from mtable 
  where userEmail != 'null' 
  Group By userEmail Order By views DESC ) AS T
WHERE T.userEmail='somename@somewhere.com'

Answer 2

在SQL中，下面的临时表检查用户可以查询相同的内容。

create table #tmp (randomno int,email varchar(10),viewsum int)
insert into #tmp select  ROW_NUMBER() over (order by sum(vi) desc) as 'rank', email,SUM(vi) 
from test
 group by email




select * from #tmp where email = 'B'
 drop table #tmp

Answer 3

我正在回答这个问题，因为@Hogan声称要采用一般方法。有一种适用于所有SQL数据库的通用方法，而不是它。

一般方法需要自我加入：

select t.useremail, t.views, count(*) as rownumber
from t left join
     t tless
     on t.useremail = tless.useremail and
        t.views >= tless.views
group by t.useremail, t.views

如果您在视图上有重复项，则以下内容相当于密集排名：

select t.useremail, t.views, count(distinct tless.views) as denserank
from t left join
     t tless
     on t.useremail = tless.useremail and
        t.views >= tless.views
group by t.useremail, t.views

并且，由于查询没有明确提及数据库，因此在大多数数据库中，您只需使用：

select t.usermail, t.views, row_number() over (partition by usermail order by views)
from t

在表中的所有用户中获得用户的等级

3 个答案: