从信用卡和借记卡记录计算余额

时间:2016-01-05 07:30:05

标签: sql database oracle oracle11g

我的数据库表中有以下记录:

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

我想计算并显示每条记录的余额,如下所示:

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

有谁可以帮助我实现上述结果?

4 个答案:

答案 0 :(得分:3)

您应该能够使用LAG分析函数来查看上一行的数据。

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

1参数表示它看起来在前一行,0参数表示如果给定偏移处的行不存在,它将返回0(即第一行)。

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

答案 1 :(得分:3)

在分析版本中生成余额使用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

如果您的表包含增加的主键,则可以使用此键而不是生成的行号。

测试数据和输出:

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

答案 2 :(得分:0)

试试这个,首先进行自我加入,然后按顺序列出结果:

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;

答案 3 :(得分:0)

对于可能会看到 mssql 的此问题的人,您可以使用此查询:

select tdate, credit, debit, 
       sum(isnull(credit, 0)-isnull(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