Sql 根据贷方和借方记录计算余额

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

---

有人能帮我得到上面的结果吗？

你应该能够使用滞后分析函数来查看前一行的数据

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

1参数意味着它看起来上一行，0参数意味着如果给定偏移量处的行不存在，它将返回0，即第一行

---

来源：

您应该能够使用滞后分析功能查看前一行的数据

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

1参数意味着它看起来上一行，0参数意味着如果给定偏移量处的行不存在，它将返回0，即第一行

---

来源：

试试这个，首先做一个自联接，然后按如下顺序列出结果：

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;

---

尝试此操作，首先执行自联接，然后按如下顺序列出结果：

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;

要生成余额，请在分析版本中使用总和

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

如果表包含递增的主键，则可以使用该主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

要生成余额，请在分析版本中使用总和

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

如果表包含递增的主键，则可以使用该主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

---

对于可能看到mssql问题的用户，您可以使用以下查询：

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

---

对于可能看到mssql问题的用户，您可以使用以下查询：

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

---

它已经在问题中发布了。第一部分显示可用数据，第二部分显示问题中已公布的预期结果。第一部分显示可用数据，第二部分显示预期结果。我不认为这是一个问题，因为学分被添加到0中。我不认为这是一个问题，因为学分被添加到0中。