sql - 按列分组并按列聚合

Question

我有一个要求，我的数据看起来像

Customer   OpenBal  Qty  Date
---------  -------  ---  --------
707001304  597      -48  20100414
707001304  597      -30  20100415
707001304  597      -30  20100419
707001304  597      -54  20100420

我想计算并获取数据

Customer   OpenBal  Qty  Date      ClosingBal (OpenBal+Qty)
---------  -------  ---  --------  ----------
707001304  597      -48  20100414  549
707001304  549      -30  20100415  519
707001304  519      -30  20100419  489
707001304  489      -54  20100420  435

可以帮忙，我如何汇总数据Qty并Date减去OpenBal一天。ClosingBal一天将打开第二天的余额。

请指教。

Answer 1

也许我遗漏了一些东西，但你不能只添加列+吗？

select  Customer
,       OpenBal
,       Qty
,       Date
,       OpenBal + Qty as ClosingBal
from    YourTable

Answer 2

原始表的每一行都包含相同的 OpenBal。OP 希望它是前一天的 CloseBal

对于 Oracle 10g、11g、SQL Server 2012 和 DB2 9.5，您可以使用 LAG 函数访问先前计算的值

Answer 3

这可能会有所帮助：

SELECT Customer, OpenBal, Date, SUM(OpenBal + Qty) AS Closingbal FROM Table_Name
GROUP BY Date, Customer, OpenBal

Answer 4

尝试这个：

insert into t_Customer values

(707001304,597,-48,20100414),
(707001304,597,-30,20100415),
(707001304,597,-30,20100419),
(707001304,597,-54,20100420)

select * from t_Customer

;WITH CTE as(select ROW_NUMBER() over (order by Customer) as sno,Customer,OpenBal,Qty,date,OpenBal+Qty as clsBal from t_Customer)
,CTE1 as (
select sno,Customer,OpenBal,Qty,date,clsBal from CTE where sno=1
union all
select c.sno,c.Customer,c1.clsBal,c.Qty,c.date,c1.clsBal+c.qty from CTE1 c1 inner join CTE c on c1.sno+1 =c.sno 
)
SELECT * FROM CTE1