如何从1行获取查询值以使用到另一行?

时间:2012-11-30 11:04:38

标签: sql database sql-server-2008 stored-procedures

这是示例查询:

payment_Type    payment_value       cost_type      cost value
Cost I          100                 Registration       40
Cost I          100                 books              40
Cost I          100                 Lab                40

COST I 有3个元素Cost_Type,它们都有自己的Cost_value

我想像这样操纵:

payment_Type    payment_value       cost_type      cost value     Payment_by_cost_type
Cost I          100                 Registration       40              40
Cost I          100                 books              40              40
Cost I          100                 Lab                40              20

重点是我想将payment_value划分为每个cost_value。在示例中,payment_by_cost变为40,40,20 = 100。

实验室cost_value为40,但它可以指定值为20,因为它来自上面划分的2成本类型。

我可以在下一行记录中使用Payment_by_cost_type中的值吗?我一直在尝试将值Payment_by_cost_type插入临时表,但select不能有insert语句。

有没有人对如何解决这个问题有任何想法?我一直在咨询DWH,他说必须使用Store程序才能通过查询完成。

4 个答案:

答案 0 :(得分:2)

我猜您的表不仅包含“Cost I”,还包含其他值,因此这里是一个查询,用于输出表格中所有组(按Payment_type)的结果:

;with table1 as
(select 
t.*,
row_number() 
 OVER 
(PARTITION BY payment_Type order by cost_type) rn
from t
)
,table2 as
( select t4.*,isnull((select sum(cost_value) from table1 
    where table1.payment_type=t4.payment_type and rn<t4.rn),0) CumSum 
  from table1 t4
) 
select payment_type,payment_value,cost_type,cost_value,
case when cost_value+CumSum<=payment_value then cost_value
  else 
    payment_value-Cumsum                  
  end                     

from table2
order by Payment_type,rn;

SQLFIDDLE demo

答案 1 :(得分:1)

您需要为记录定义某种订单,以定义付款的应用顺序

完成后(我在这个例子中使用ID)......

select *
    , case
          when payment_value-(select isnull(SUM(cost_value),0) from yourtable t2 where t2.id<t1.id)<cost_value
          then payment_value-(select isnull(SUM(cost_value),0) from yourtable t2 where t2.id<t1.id)
          else cost_value
          end
from yourtable t1

答案 2 :(得分:1)

使用常用表表达式逐步执行。

declare @t table (
payment_type varchar(20),
payment_value int,
cost_type varchar(20),
cost_value int,
cost_id int --for the ordering
)

insert @t values 
('Cost I',100,'Registration',40,1),
('Cost I',100,'books',40,2),
('Cost I',100,'Lab',40,3),
('Cost 2',100,'Registration',40,4),
('Cost 2',100,'books',50,5),
('Cost 2',100,'Lab',40,6)

--get count for each payment_type to determine last row
;with payment_value_cte(payment_type,payment_value,count) as
(
    select payment_type,payment_value,COUNT(*) from @t group by payment_type,payment_value
),
--use sequential index for each row in payment type
payment_value_index_cte(
    payment_type ,
    payment_value,
    cost_type,
    cost_value,
    cost_id,
    row) as
(
    select *,ROW_NUMBER() OVER(PARTITION BY payment_type ORDER BY cost_id) from @t --assumes order is by an id
),
--get sum of each row for payment type except last row
payment_value_sum_except_last_cte(
    payment_type,
    payment_value,
    current_sum) as
(
    select pi.payment_type,pi.payment_value,SUM(cost_value) 
    from payment_value_index_cte pi
    inner join payment_value_cte pt on pt.payment_type = pi.payment_type
    where pi.row < pt.count
    group by pi.payment_type,pi.payment_value
)
select 
pi.payment_type,pi.payment_value,pi.cost_type,pi.cost_value,
--if last row calculate difference, else use the cost_value
case when pi.row = pt.count then pt.payment_value - pe.current_sum else pi.cost_value end [Payment_by_cost_type]   
from payment_value_index_cte pi
inner join payment_value_cte pt on pt.payment_type = pi.payment_type
inner join payment_value_sum_except_last_cte pe on pe.payment_type = pi.payment_type

答案 3 :(得分:1)

SELECT payment_Type, payment_value, cost_type, cost_value,
       CASE WHEN ROW_NUMBER() OVER (ORDER BY(SELECT 1)) = COUNT(*) OVER (PARTITION BY payment_Type)
            THEN SUM(cost_value) OVER (PARTITION BY payment_Type) - payment_value 
            ELSE cost_value END AS Payment_by_cost_type
FROM dbo.your_table

SQLFiddle上的演示