我希望我能以一种可以理解的方式描述我的挑战.我在Oracle Database 12c上有两个表,如下所示:
表名"发票"
I_ID | invoice_number | creation_date | i_amount
------------------------------------------------------
1 | 10000000000 | 01.02.2016 00:00:00 | 30
2 | 10000000001 | 01.03.2016 00:00:00 | 25
3 | 10000000002 | 01.04.2016 00:00:00 | 13
4 | 10000000003 | 01.05.2016 00:00:00 | 18
5 | 10000000004 | 01.06.2016 00:00:00 | 12
表名"付款"
P_ID | reference | received_date | p_amount
------------------------------------------------------
1 | PAYMENT01 | 12.02.2016 13:14:12 | 12
2 | PAYMENT02 | 12.02.2016 15:24:21 | 28
3 | PAYMENT03 | 08.03.2016 23:12:00 | 2
4 | PAYMENT04 | 23.03.2016 12:32:13 | 30
5 | PAYMENT05 | 12.06.2016 00:00:00 | 15
所以我想要一个select语句(可能有oracle分析函数但我并不熟悉它),其中的付款总结到达到发票金额,按日期排序.如果两笔付款的总和超过发票金额,则最后一笔付款金额的其余部分应用于下一张发票.
在此示例中,结果应如下所示:
invoice_number | reference | used_pay_amount | open_inv_amount
----------------------------------------------------------
10000000000 | PAYMENT01 | 12 | 18
10000000000 | PAYMENT02 | 18 | 0
10000000001 | PAYMENT02 | 10 | 15
10000000001 | PAYMENT03 | 2 | 13
10000000001 | PAYMENT04 | 13 | 0
10000000002 | PAYMENT04 | 13 | 0
10000000003 | PAYMENT04 | 4 | 14
10000000003 | PAYMENT05 | 14 | 0
10000000004 | PAYMENT05 | 1 | 11
如果有一个带有"简单"select语句的解决方案,那就太好了.
提前为你的时间...
Oracle安装程序:
CREATE TABLE invoices ( i_id, invoice_number, creation_date, i_amount ) AS
SELECT 1, 100000000, DATE '2016-01-01', 30 FROM DUAL UNION ALL
SELECT 2, 100000001, DATE '2016-02-01', 25 FROM DUAL UNION ALL
SELECT 3, 100000002, DATE '2016-03-01', 13 FROM DUAL UNION ALL
SELECT 4, 100000003, DATE '2016-04-01', 18 FROM DUAL UNION ALL
SELECT 5, 100000004, DATE '2016-05-01', 12 FROM DUAL;
CREATE TABLE payments ( p_id, reference, received_date, p_amount ) AS
SELECT 1, 'PAYMENT01', DATE '2016-01-12', 12 FROM DUAL UNION ALL
SELECT 2, 'PAYMENT02', DATE '2016-01-13', 28 FROM DUAL UNION ALL
SELECT 3, 'PAYMENT03', DATE '2016-02-08', 2 FROM DUAL UNION ALL
SELECT 4, 'PAYMENT04', DATE '2016-02-23', 30 FROM DUAL UNION ALL
SELECT 5, 'PAYMENT05', DATE '2016-05-12', 15 FROM DUAL;
查询:
WITH total_invoices ( i_id, invoice_number, creation_date, i_amount, i_total ) AS (
SELECT i.*,
SUM( i_amount ) OVER ( ORDER BY creation_date, i_id )
FROM invoices i
),
total_payments ( p_id, reference, received_date, p_amount, p_total ) AS (
SELECT p.*,
SUM( p_amount ) OVER ( ORDER BY received_date, p_id )
FROM payments p
)
SELECT invoice_number,
reference,
LEAST( p_total, i_total )
- GREATEST( p_total - p_amount, i_total - i_amount ) AS used_pay_amount,
GREATEST( i_total - p_total, 0 ) AS open_inv_amount
FROM total_invoices
INNER JOIN
total_payments
ON ( i_total - i_amount < p_total
AND i_total > p_total - p_amount );
说明:
两个子查询factoring(WITH ... AS ())子句只是向invoices和payments表添加一个额外的虚拟列,其中包含发票/付款金额的累计总和.
您可以将范围与每张发票(或付款)相关联,作为发票(付款)之前的欠款(已付)累计金额以及之后的欠款(已付款)累计金额.然后可以在这些范围重叠的地方连接这两个表.
这open_inv_amount是累计发票金额与累计支付金额之间的正差额.
在used_pay_amount稍微复杂一些,但你需要找到当前累计发票和付款总额的低,以前的累积发票和付款总额较高的区别.
输出:
INVOICE_NUMBER REFERENCE USED_PAY_AMOUNT OPEN_INV_AMOUNT
-------------- --------- --------------- ---------------
100000000 PAYMENT01 12 18
100000000 PAYMENT02 18 0
100000001 PAYMENT02 10 15
100000001 PAYMENT03 2 13
100000001 PAYMENT04 13 0
100000002 PAYMENT04 13 0
100000003 PAYMENT04 4 14
100000003 PAYMENT05 14 0
100000004 PAYMENT05 1 11
更新:
基于mathguy使用UNION加入数据的方法,我想出了一个不同的解决方案,重新使用了我的一些代码.
WITH combined ( invoice_number, reference, i_amt, i_total, p_amt, p_total, total ) AS (
SELECT invoice_number,
NULL,
i_amount,
SUM( i_amount ) OVER ( ORDER BY creation_date, i_id ),
NULL,
NULL,
SUM( i_amount ) OVER ( ORDER BY creation_date, i_id )
FROM invoices
UNION ALL
SELECT NULL,
reference,
NULL,
NULL,
p_amount,
SUM( p_amount ) OVER ( ORDER BY received_date, p_id ),
SUM( p_amount ) OVER ( ORDER BY received_date, p_id )
FROM payments
ORDER BY 7,
2 NULLS LAST,
1 NULLS LAST
),
filled ( invoice_number, reference, i_prev, i_total, p_prev, p_total ) AS (
SELECT FIRST_VALUE( invoice_number ) IGNORE NULLS OVER ( ORDER BY ROWNUM ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING ),
FIRST_VALUE( reference ) IGNORE NULLS OVER ( ORDER BY ROWNUM ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING ),
FIRST_VALUE( i_total - i_amt ) IGNORE NULLS OVER ( ORDER BY ROWNUM ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING ),
FIRST_VALUE( i_total ) IGNORE NULLS OVER ( ORDER BY ROWNUM ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING ),
FIRST_VALUE( p_total - p_amt ) IGNORE NULLS OVER ( ORDER BY ROWNUM ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING ),
COALESCE(
p_total,
LEAD( p_total ) IGNORE NULLS OVER ( ORDER BY ROWNUM ),
LAG( p_total ) IGNORE NULLS OVER ( ORDER BY ROWNUM )
)
FROM combined
),
vals ( invoice_number, reference, upa, oia, prev_invoice ) AS (
SELECT invoice_number,
reference,
COALESCE( LEAST( p_total - i_total ) - GREATEST( p_prev, i_prev ), 0 ),
GREATEST( i_total - p_total, 0 ),
LAG( invoice_number ) OVER ( ORDER BY ROWNUM )
FROM filled
)
SELECT invoice_number,
reference,
upa AS used_pay_amount,
oia AS open_inv_amount
FROM vals
WHERE upa > 0
OR ( reference IS NULL AND invoice_number <> prev_invoice AND oia > 0 );
说明:
该combined子查询保条款加入了两个表UNION ALL,并生成发票和支付的金额累计总量.它做的最后一件事就是按行的递增累计总数对行进行排序(如果存在联系,它将按照创建的顺序将付款放在发票之前).
该filled子查询保条款将填补之前生成的表,这样,如果一个值为null,则它会从下一个非空行的值(如果有没有付款的发票这时就会发现总的先前行的先前付款).
该vals子查询保条款适用相同的计算,我以前的查询(见上文).它还添加了该prev_invoice列,以帮助识别完全未付的发票.
最终SELECT获取值并过滤掉不必要的行.
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