Regex 正则表达式以排除大括号之间的值，并对其余部分应用模式

我在一个字段中有以下值

Aa11
BBB-
BBB+
A- /*-
A3
Ca

我会用正则表达式

(([A-Z](([abc]+\d?)|\d))|([A-Z]+[+-]?)

效果很好。然而，现在我有了另一组新的数据

(p)A3
(q)A- /*-

如何确保忽略括号和括号之间的值以应用上述正则表达式

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我在oracle中使用REGEX_SUBSTR来实现这一点。

正则表达式

\（.*？）

将匹配开头和结尾括号，括号之间的字符尽可能少，而

[^（]*？

将匹配零个或尽可能少的非开头括号字符。您可以将这些字符组合起来，得到正则表达式

^（[^（]*\（.*？\）*？[^（]*？

这将匹配尽可能少的括号组（前提是您没有嵌套的括号），直到找到所需的模式

Oracle 11g R2架构设置：

CREATE TABLE data ( value ) AS
SELECT 'Aa11' FROM DUAL UNION ALL
SELECT 'BBB-' FROM DUAL UNION ALL
SELECT 'BBB+' FROM DUAL UNION ALL
SELECT 'A- /*-' FROM DUAL UNION ALL
SELECT 'A3' FROM DUAL UNION ALL
SELECT 'Ca' FROM DUAL UNION ALL
SELECT '(p)A3' FROM DUAL UNION ALL
SELECT '(q)A- /*-' FROM DUAL UNION ALL
SELECT '(Ca)Cb(Cc)' FROM DUAL UNION ALL
SELECT '--(Ca)--(Cb)--Cc(--Ca)' FROM DUAL;

SELECT value,
       REGEXP_SUBSTR(
         value,
         '^([^(]*?\(.*?\))*?[^(]*?([A-Z]([abc]+\d?|\d|[A-Z]*[+-]?))',
         1,    -- Start at 1st character
         1,    -- Find the 1st occurrence
         NULL, -- No flags
         2     -- Return 2nd capturing group
       ) AS regex_output
FROM   data

|                  VALUE | REGEX_OUTPUT |
|------------------------|--------------|
|                   Aa11 |          Aa1 |
|                   BBB- |         BBB- |
|                   BBB+ |         BBB+ |
|                 A- /*- |           A- |
|                     A3 |           A3 |
|                     Ca |           Ca |
|                  (p)A3 |           A3 |
|              (q)A- /*- |           A- |
|             (Ca)Cb(Cc) |           Cb |
| --(Ca)--(Cb)--Cc(--Ca) |           Cc |

查询1：

CREATE TABLE data ( value ) AS
SELECT 'Aa11' FROM DUAL UNION ALL
SELECT 'BBB-' FROM DUAL UNION ALL
SELECT 'BBB+' FROM DUAL UNION ALL
SELECT 'A- /*-' FROM DUAL UNION ALL
SELECT 'A3' FROM DUAL UNION ALL
SELECT 'Ca' FROM DUAL UNION ALL
SELECT '(p)A3' FROM DUAL UNION ALL
SELECT '(q)A- /*-' FROM DUAL UNION ALL
SELECT '(Ca)Cb(Cc)' FROM DUAL UNION ALL
SELECT '--(Ca)--(Cb)--Cc(--Ca)' FROM DUAL;

SELECT value,
       REGEXP_SUBSTR(
         value,
         '^([^(]*?\(.*?\))*?[^(]*?([A-Z]([abc]+\d?|\d|[A-Z]*[+-]?))',
         1,    -- Start at 1st character
         1,    -- Find the 1st occurrence
         NULL, -- No flags
         2     -- Return 2nd capturing group
       ) AS regex_output
FROM   data

|                  VALUE | REGEX_OUTPUT |
|------------------------|--------------|
|                   Aa11 |          Aa1 |
|                   BBB- |         BBB- |
|                   BBB+ |         BBB+ |
|                 A- /*- |           A- |
|                     A3 |           A3 |
|                     Ca |           Ca |
|                  (p)A3 |           A3 |
|              (q)A- /*- |           A- |
|             (Ca)Cb(Cc) |           Cb |
| --(Ca)--(Cb)--Cc(--Ca) |           Cc |

：

CREATE TABLE data ( value ) AS
SELECT 'Aa11' FROM DUAL UNION ALL
SELECT 'BBB-' FROM DUAL UNION ALL
SELECT 'BBB+' FROM DUAL UNION ALL
SELECT 'A- /*-' FROM DUAL UNION ALL
SELECT 'A3' FROM DUAL UNION ALL
SELECT 'Ca' FROM DUAL UNION ALL
SELECT '(p)A3' FROM DUAL UNION ALL
SELECT '(q)A- /*-' FROM DUAL UNION ALL
SELECT '(Ca)Cb(Cc)' FROM DUAL UNION ALL
SELECT '--(Ca)--(Cb)--Cc(--Ca)' FROM DUAL;

SELECT value,
       REGEXP_SUBSTR(
         value,
         '^([^(]*?\(.*?\))*?[^(]*?([A-Z]([abc]+\d?|\d|[A-Z]*[+-]?))',
         1,    -- Start at 1st character
         1,    -- Find the 1st occurrence
         NULL, -- No flags
         2     -- Return 2nd capturing group
       ) AS regex_output
FROM   data

|                  VALUE | REGEX_OUTPUT |
|------------------------|--------------|
|                   Aa11 |          Aa1 |
|                   BBB- |         BBB- |
|                   BBB+ |         BBB+ |
|                 A- /*- |           A- |
|                     A3 |           A3 |
|                     Ca |           Ca |
|                  (p)A3 |           A3 |
|              (q)A- /*- |           A- |
|             (Ca)Cb(Cc) |           Cb |
| --(Ca)--(Cb)--Cc(--Ca) |           Cc |

---

您的正则表达式无效。请发布有效的正则表达式，并解释确切的问题是什么。鉴于上述输入，您最终需要得到什么？您可以尝试使用：

（\（[^）]*\）捕捉参数？

忽略此子项谢谢，这正是我所需要的。不过，我仍然了解此解决方案的工作原理。我已将其分解为两个步骤，删除任何带大括号的内容，然后应用较旧的正则表达式提取值