2

我有以下数据结构

Customer | Order | Date       | Amount | TransactionType
AABB     | AB01  | 2012-06-01 | 3000   | Invoiced
AABB     | AB01  | 2012-06-05 | 3000   | Payment

我需要从发票提交之日到付款之日重复我的数据。像这样;

Customer|Order|Date      | AmountDue|AmountPaid|DatePaid  |TransactionType
AABB    |AB01 |2012-06-01| 3000     |NULL      |2012-06-05|Invoiced
AABB    |AB01 |2012-06-02| 3000     |NULL      |2012-06-05|Invoiced
AABB    |AB01 |2012-06-03| 3000     |NULL      |2012-06-05|Invoiced
AABB    |AB01 |2012-06-04| 3000     |NULL      |2012-06-05|Invoiced
AABB    |AB01 |2012-06-05| 3000     |3000      |2012-06-05|Payment

如何为这种情况创建 SQL 脚本

4

5 回答 5

2

尝试这个：

  create table cust1
       (
          Customer varchar(20),
          Orders varchar(10),
          Date datetime,
          Amount float,
          TransactionType varchar(50)
        )
        INSERT INTO cust1
        VALUES('AABB','AB01','2012-06-01',3000,'Invoiced'),
        ('AABB','AB01','2012-06-05',3000,'Payment')

           DECLARE @stDate datetime,@eddate datetime
select @stDate =MIN(date),@eddate =MAX(date) from cust1

select c1.Customer,c1.Orders,DATEADD(DD,number,@stDate) as [date],
amount amountDue,
CASE WHEN (DATEADD(DD,number,@stDate)) = @eddate then amount else null end as amountPaid,
@eddate as datepaid,
CASE WHEN (DATEADD(DD,number,@stDate)) <> @eddate then 'Invoiced' else 'Payment' end as TransactionType

from master..spt_values p inner join cust1 c1
on right(cast(c1.date as DATE),2) <= (case when p.number = 0 then 1 else p.number end)
where type='p'and DATEADD(DD,number,@stDate) <=@eddate
于 2012-08-02T15:37:05.083 回答
1

这是一个尝试，假设有两行，已付款，发票行和付款行都存在，并且您正在尝试解决一个特定的客户/订单。

DECLARE @t TABLE
(
  Customer VARCHAR(32), 
  [Order] VARCHAR(32), 
  [Date] DATE, 
  Amount INT, 
  TransactionType VARCHAR(32)
);

INSERT @t VALUES
('AABB','AB01','2012-06-01',3000,'Invoiced'),
('AABB','AB01','2012-06-05',3000,'Payment');


;WITH t AS (SELECT * FROM @t AS t WHERE t.Customer = 'AABB' AND t.[Order] = 'AB01'),
rng AS (SELECT s = MIN([Date]), e = MAX([Date]) FROM t), 
n AS (SELECT TOP (DATEDIFF(DAY, (SELECT s FROM rng), (SELECT e FROM rng))) 
      n = ROW_NUMBER() OVER (ORDER BY [object_id]) - 1 FROM sys.all_objects
)
SELECT t.Customer, t.[Order], [Date] = DATEADD(DAY, n.n, t.[Date]), 
    AmountDue = Amount, AmountPaid = NULL, DatePaid = (SELECT e FROM rng), 
    t.TransactionType
FROM n CROSS JOIN t WHERE t.[Date] < (SELECT e FROM rng)
UNION ALL 
SELECT t.Customer, t.[Order], t.[Date], 
    AmountDue = NULL, AmountPaid = Amount, DatePaid = t.[Date], 
    t.TransactionType
FROM t WHERE t.[Date] = (SELECT e FROM rng);
于 2012-08-02T15:57:32.610 回答
0

这是一个通用 SQL 版本，适用于没有一些更高级功能的 RDBMS。还具有列出尚未付款的帐户的额外好处。它确实假设您有一个可用的日历文件（真实的或临时的）

SELECT history.customer, history.order, calendar.calendar_date,
       CASE WHEN history.transaction_type = 'Invoiced'
            THEN history.amount END as amount_due,
       CASE WHEN history.transaction_type = 'Payment'
            THEN history.amount END as amount_paid,
       history.transaction_type
FROM Transaction_History as history
JOIN Calendar
ON calendar.calendar_date < CURRENT_DATE + 1 DAY
AND ((calendar.calendar_date >= history.date 
      AND history.transaction_type = 'Invoiced')
     OR (calendar.calendar_date = history.date
         AND history.transaction_type = 'Payment'))
LEFT JOIN Transaction_History as exclusion
ON exclusion.customer = history.customer
AND exclusion.order = history.order
AND exclusion.transaction_type = 'Payment'
AND history.transaction_type = 'Invoiced'
AND exclusion.date <= calendar.calendar_date
WHERE exclusion.customer IS NULL

在我的系统 (DB2) 上测试 - SQLFiddle 似乎已关闭。

于 2012-08-02T16:20:16.310 回答
0

好的，我知道 OP 已经被选择了一个解决方案。但是我用行生成器发布了我自己的解决方案。我认为这是一种优雅的方式来做到这一点。未来在这里：

拳头，创建表格并填充值：

create table #t (
Customer char(4), [Order] char(4), [Date] date, Amount money,
TransactionType varchar(50)
)
insert into #t values 
( 'AABB','AB01','2012-06-01',3000,'Invoiced'),
( 'AABB','AB01','2012-06-05',3000,'Payment');

在这里，行生成器和它自己的查询：

declare @fromdate date
set @fromdate = '1/1/2001'
;WITH 
Nbrs_3( n ) AS ( SELECT 1 UNION SELECT 0 ),
Nbrs_2( n ) AS ( SELECT 1 FROM Nbrs_3 n1 CROSS JOIN Nbrs_3 n2 ),
Nbrs_1( n ) AS ( SELECT 1 FROM Nbrs_2 n1 CROSS JOIN Nbrs_2 n2 ),
Nbrs_0( n ) AS ( SELECT 1 FROM Nbrs_1 n1 CROSS JOIN Nbrs_1 n2 ),
Nbrs ( n ) AS ( SELECT 1 FROM Nbrs_0 n1 CROSS JOIN Nbrs_0 n2 ),
all_dates as (
  SELECT 
     dateadd( day, n , @fromDate) as [date]
  FROM ( SELECT ROW_NUMBER() OVER (ORDER BY n)
  FROM Nbrs ) D ( n )
),
all_intervals as (
  select 
     t1.customer, t1.[order], t1.[date] as date_from, t1.amount as amount_due,
     t2.[date] as date_paid
  from
     #t t1 inner join
     #t t2 on t2.TransactionType = 'Payment'
              and t1.customer = t2.customer and t1.[order] = t2.[order]
  where t1.TransactionType = 'Invoiced'
  )
select a.*, d.[date]
from all_intervals a
inner join all_dates d 
   on d.[date] between a.date_from and a.date_paid

结果：

customer order date_from     amount_due date_paid     date          
-------- ----- ------------- ---------- ------------- ------------- 
AABB     AB01  2012-06-01 00:00:003000       2012-06-05 00:00:002012-06-01 00:00:00
AABB     AB01  2012-06-01 00:00:003000       2012-06-05 00:00:002012-06-02 00:00:00
AABB     AB01  2012-06-01 00:00:003000       2012-06-05 00:00:002012-06-03 00:00:00
AABB     AB01  2012-06-01 00:00:003000       2012-06-05 00:00:002012-06-04 00:00:00
AABB     AB01  2012-06-01 00:00:003000       2012-06-05 00:00:002012-06-05 00:00:00

你可以试试。

于 2012-08-02T16:10:40.660 回答
0

首先，您需要汇总获得发票日期和付款日期所需的数据。然后，您可以为其他记录生成序列。一种方法是使用递归 CTE。但是，我通常会执行以下操作：

客户 | 订购 | 日期 | 金额 | 交易类型

select Customer, Order,
       dateadd(d seqnum, InvoiceDate) as Date,
       InvoiceAmount as AmountDue,
       PaymentAmount as AmountPaid,
       PaymentDate as DatePaid,
       (case when dateadd(d seqnum, InvoiceDate) <> PaymentDate then 'Invoice' else 'Payment' end) as TransactionType
from (select customer, order,
             max(case when TransactionType = 'Invoice' then amount end) as InvoiceAmount,
             max(case when TransactionType = 'Payment' then amountend) as PaymentAmount
             max(case when TransactionType = 'Invoice' then date end) as InvoiceDate,
             max(case when TransactionType = 'Payment' then date end) as PaymentDate
      from t
      group by customer, order
     ) t join
     (select row_number() over (order by (select NULL)) - 1 as seqnum
      from INFORMATION_SCHEMA.Columns
     ) seq
     on dateadd(d, 1, InvoiceDate, seqnum) <= PaymentDate

这是假设您有一个发票记录和一个付款记录。它还假设天数不是那么大。InformationSchema.Columns 的使用只是为了生成一个序列。

于 2012-08-02T15:40:40.323 回答