计数sql行，将每一行的值添加到下一行

Question

我有两列，我想将每一行的值添加到下一行。 我有两列4行表是这样的：

date   users 
2011-01-01  1 
2011-02-02  1 
2011-03-02  2 
2011-04-02  4 

，我希望用户显示如下图所示，并添加每一行到它的上一行值：

用户：

1 
2 
4 
8 

是否有一个MS SQL Server的功能呢？

感谢

Answer 1

在SQL Server 2 005及以上（您没有指定您的版本），您可以使用两个CTE（公用表表达式）做这样的事情 - 第一个叫做UserSequence的CTE将数据放入订单并给它一个序号（Sequence ），以及第二递归CTE（公共表表达式）来计算运行总计：

;WITH UserSequence AS 
(
    SELECT 
     Date, Users, ROW_NUMBER() OVER(ORDER BY Date) as 'Sequence' 
    FROM 
     dbo.YourTable 
), 
UserValues AS 
(
    SELECT 
    u.Users AS 'UserValue', u.Date, u.Sequence 
    FROM UserSequence u 
    WHERE Sequence = 1 

    UNION ALL 

    SELECT 
    u.Users + uv.UserValue AS 'UserValue', u.Date, u.Sequence 
    FROM UserSequence u 
    INNER JOIN UserValues uv ON u.Sequence = uv.Sequence + 1 
) 
SELECT 
    Date, Sequence, UserValue AS 'Users' 
FROM 
    UserValues 
ORDER BY 
    Sequence 

这应该给你的输出是这样的：

Date      Sequence Users 
2011-01-01 00:00:00.000  1   1 
2011-02-02 00:00:00.000  2   2 
2011-03-02 00:00:00.000  3   4 
2011-04-02 00:00:00.000  4   8 

Answer 2

这里是你如何能做到这一点：

set @total_users = 0; 
select users, @total_users:=@total_users + users as total_users from tablename; 

Answer 3

这是ANSI SQL的解决方案：

 
SELECT users + lag(users) over (order by users) 
FROM your_table 
ORDER BY users 

作品在PostgreSQL里，甲骨文，DB2和Teradata（SQL服务器确实有窗口函数，但不幸的是不是lag（）函数）

Answer 4

With cte_Test(Sequence,Name,Value) 
As 
(
select ROW_NUMBER() OVER(ORDER BY value) as 'Sequence' ,name,value from test 
) 

select t1.Name,sum(T2.Value) AS Value from cte_Test t1 
cross JOIN cte_Test t2 where t2.Sequence <= t1.Sequence 
group BY t1.Name 
order by 2 

Answer 5

这是可能通过交叉加入。如果您的表格列处于Asc顺序。

select <T1.Text>,sum(T2.Value) AS 'Commulative Value' from <Your Table name>t1 
cross JOIN <Your Table name> t2 where t2.value<= t1.value 
group BY t1.text 
order by value