Sql 使用以前的值计算字段

我面临一个问题，涉及到计算字段的值，但使用前一行的值。表中包含数据的速率

ID Original Rate      Capital
-- ------------------ -------
 1 50                       1
 2 50 + 1                   1
 3 50 + 1 + 1               1
 4 50 + 1 + 1 + 2           2
 5 50 + 1 + 1 + 2 + 2       2

公式为：原始利率（n）=原始利率（n-1）+资本

在original中，我有1行原始利率为50，我希望下一个值应该是原始利率+资本

我也尝试过使用，但它似乎不适用于计算字段

如果你们对这个案子有任何想法，请帮帮我

谢谢你抽出时间

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新案例的更新：

+----+---------------+------+---------+-------+-------+-------+
| ID | Original Rate | Type | Capital | Rate1 | Rate2 | Rate3 |
+----+---------------+------+---------+-------+-------+-------+ 
|  1 |            50 |    1 |       2 | 0.1   |       |       | 
|  2 |            50 |    1 |       2 | 0.1   |       |       |
|  3 |            50 |    1 |       2 | 0.1   |       |       |
|  4 |            50 |    2 |       1 |       | 0.2   |       |
|  5 |            50 |    3 |       5 |       |       | 0.3   |
+----+---------------+------+---------+-------+-------+-------+

New Rate (n) = New Rate (n-1) + Original Rate * IF(Type=1,Rate1*capital,IF(Type=2,Rate2*capital),Rate3*capital)  

我期望的输出在最后一列中，如下所示：

+----+---------------+------+---------+-------+-------+-------+-----+
| ID | Original Rate | Type | Capital | Rate1 | Rate2 | Rate3 |     |
+----+---------------+------+---------+-------+-------+-------+-----+
|  1 |            50 |    1 |       2 | 0.1   |       |       |  10 |
|  2 |            50 |    1 |       2 | 0.1   |       |       |  20 |
|  3 |            50 |    1 |       2 | 0.1   |       |       |  30 |
|  4 |            50 |    2 |       4 |       | 0.2   |       |  70 |
|  5 |            50 |    3 |       5 |       |       | 0.3   | 145 |
+----+---------------+------+---------+-------+-------+-------+-----+

欢迎任何意见和建议

[2017年4月17日更新] 我使用递归和

WITH RATE_MASTER AS (ID, RATE)
(
    SELECT 1, ORIGINALRATE FROM RATE WHERE ID = 1
    
    UNION ALL
    
    SELECT ID,
    CASE 
        WHEN TYPE = 1 THEN RATE + Rate1*Capital 
        WHEN TYPE = 2 THEN RATE + Rate2*Capital
        WHEN TYPE = 3 THEN RATE + Rate3*Capital
    END AS RATE
    
    FROM RATE, RATE_MASTER
    WHERE RATE_MASTER.ID = RATE.ID - 1
)
SELECT * FROM RATE_MASTER;

它工作正常

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[关闭]Oracle安装程序：

CREATE TABLE rate ( id, capital ) AS
  SELECT 1, 1 FROM DUAL UNION ALL
  SELECT 2, 1 FROM DUAL UNION ALL
  SELECT 3, 1 FROM DUAL UNION ALL
  SELECT 4, 2 FROM DUAL UNION ALL
  SELECT 5, 2 FROM DUAL;

SELECT id,
       50 + SUM( capital ) OVER ( ORDER BY id )
          - FIRST_VALUE( capital ) OVER ( ORDER BY id ) AS rate,
       capital
FROM   rate;

ID RATE CAPITAL
-- ---- -------
 1   50       1
 2   51       1
 3   52       1
 4   54       2
 5   56       2

查询：

CREATE TABLE rate ( id, capital ) AS
  SELECT 1, 1 FROM DUAL UNION ALL
  SELECT 2, 1 FROM DUAL UNION ALL
  SELECT 3, 1 FROM DUAL UNION ALL
  SELECT 4, 2 FROM DUAL UNION ALL
  SELECT 5, 2 FROM DUAL;

SELECT id,
       50 + SUM( capital ) OVER ( ORDER BY id )
          - FIRST_VALUE( capital ) OVER ( ORDER BY id ) AS rate,
       capital
FROM   rate;

ID RATE CAPITAL
-- ---- -------
 1   50       1
 2   51       1
 3   52       1
 4   54       2
 5   56       2

输出：

CREATE TABLE rate ( id, capital ) AS
  SELECT 1, 1 FROM DUAL UNION ALL
  SELECT 2, 1 FROM DUAL UNION ALL
  SELECT 3, 1 FROM DUAL UNION ALL
  SELECT 4, 2 FROM DUAL UNION ALL
  SELECT 5, 2 FROM DUAL;

SELECT id,
       50 + SUM( capital ) OVER ( ORDER BY id )
          - FIRST_VALUE( capital ) OVER ( ORDER BY id ) AS rate,
       capital
FROM   rate;

ID RATE CAPITAL
-- ---- -------
 1   50       1
 2   51       1
 3   52       1
 4   54       2
 5   56       2

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请将一些示例数据和预期结果以表格格式作为格式化文本发布。请单击“运行代码段”按钮？我还用数据更新了一个表。谢谢MT0，它可以处理示例数据。我要申请我的案子，很快就回来！