計算平衡

Question

我在我的數據庫表中的以下記錄：

Date   Credit Debit  Description 
--------------- ------- ------- --------------- 
12-24-2015  5     Purchased credit 
12-20-2015    1   Consumed credit 
12-15-2015  3     Purchased credit 
12-08-2015    1   Consumed credit 
12-08-2015    1   Consumed credit 
12-07-2015    1   Consumed credit 
12-04-2015    1   Consumed credit 
12-03-2015    1   Consumed credit 
12-01-2015  5     Purchased credit 

我想要計算並顯示每個記錄的餘額如下圖所示：

Date   Credit Debit Balance Description 
------------ ------- ------- ------- --------------- 
12-24-2015 5  0  7  Purchased credit 
12-20-2015   1  2  Consumed credit 
12-15-2015 3  0  3  Purchased credit 
12-08-2015   1  0  Consumed credit 
12-08-2015   1  1  Consumed credit 
12-07-2015   1  2  Consumed credit 
12-04-2015   1  3  Consumed credit 
12-03-2015   1  4  Consumed credit 
12-01-2015 5  0  5  Purchased credit 

能任何人都可以幫助我實現上述結果？

Answer 1

爲了產生分析天平版使用sum()。

select tdate, credit, debit, 
     sum(nvl(credit, 0)-nvl(debit, 0)) over (order by rn) balance, description 
    from (
    select tdate, credit, debit, row_number() over (order by tdate) rn, description 
     from test) 
    order by rn desc 

如果您的表格包含增加的主鍵，則可以使用此代替生成的行號。

測試數據和輸出：

create table test (tdate date, credit number(6), debit number(6), description varchar2(20)); 
insert into test values (date '2015-12-24', 5, null, 'Purchased credit'); 
insert into test values (date '2015-12-20', null, 1, 'Consumed credit'); 
insert into test values (date '2015-12-15', 3, null, 'Purchased credit'); 
insert into test values (date '2015-12-08', null, 1, 'Consumed credit'); 
insert into test values (date '2015-12-08', null, 1, 'Consumed credit'); 
insert into test values (date '2015-12-07', null, 1, 'Consumed credit'); 
insert into test values (date '2015-12-04', null, 1, 'Consumed credit'); 
insert into test values (date '2015-12-03', null, 1, 'Consumed credit'); 
insert into test values (date '2015-12-01', 5, null, 'Purchased credit'); 

TDATE  CREDIT DEBIT BALANCE DESCRIPTION 
----------- ------- ------- ---------- -------------------- 
2015-12-24  5     7 Purchased credit 
2015-12-20    1   2 Consumed credit 
2015-12-15  3     3 Purchased credit 
2015-12-08    1   0 Consumed credit 
2015-12-08    1   1 Consumed credit 
2015-12-07    1   2 Consumed credit 
2015-12-04    1   3 Consumed credit 
2015-12-03    1   4 Consumed credit 
2015-12-01  5     5 Purchased credit 

Answer 2

您應該可以使用分析函數LAG來查看上一行的數據。

SELECT Date, 
     Credit, 
     Debit, 
     LAG(Balance, 1, 0) OVER(ORDER BY Date) - Debit + Credit AS Balance, 
     Description 
    FROM sometable 

的1參數意味着它看起來1行前，所述0參數意味着如果該行的給定偏移量不存在，則返回0，而不是（即，用於第一行）。

來源：https://oracle-base.com/articles/misc/lag-lead-analytic-functions

Answer 3

爲了試試這個，首先做一個自我加入，然後列出了結果：

SELECT a1.Date, a1.Debit, a1.Credit, SUM(a2.Debit-a2.Credit) Balance, Description 
FROM table a1, table a2 
WHERE a1.debit<=a2.debit and a1.credit<=a2.credit OR (a1.debit=a2.debit and a1.credit=a2.credit and a1.date = a2.date) 
GROUP BY a1.Date, a1.Debit, a1.Credit 
ORDER BY a1.Date DESC;