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regex - 在模式匹配正则表达式中使用“或”

问题描述

我正在使用正则表达式来查找匹配的模式。但不知何故，我无法找到所有的 Occrences。

我需要匹配模式的输入文件（请注意，这是一个示例文件，实际上只有 3 次出现 - 它有多个出现）：

aaa-233- hi, how are you? 
aaa-234- 6(-8989) 
aaa-235- 123
end

所以，我希望我的输出是

 hi, how are you? 
 6(-8988) 
 123

我的正则表达式是

aaa\\-[A-Za-z0-9,->#]\\-(.+?)(aaa)

伪代码

Output= matcher.group(2);

如何使逻辑从 aaa 开始读取并结束它遇到 aaa 或结束。

标签： regex

解决方案

采用

(?sm)^aaa-[^-]+-.*?(?=\naaa|\nend|\z)

见证明

解释

                         EXPLANATION
--------------------------------------------------------------------------------
  (?ms)                    set flags for this block (with ^ and $
                           matching start and end of line) (with .
                           matching \n) (case-sensitive) (matching
                           whitespace and # normally)
--------------------------------------------------------------------------------
  ^                        the beginning of a "line"
--------------------------------------------------------------------------------
  aaa-                     'aaa-'
--------------------------------------------------------------------------------
  [^-]+                    any character except: '-' (1 or more times
                           (matching the most amount possible))
--------------------------------------------------------------------------------
  -                        '-'
--------------------------------------------------------------------------------
  .*?                      any character (0 or more times (matching
                           the least amount possible))
--------------------------------------------------------------------------------
  (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
    \n                       '\n' (newline)
--------------------------------------------------------------------------------
    aaa                      'aaa'
--------------------------------------------------------------------------------
   |                        OR
--------------------------------------------------------------------------------
    \n                       '\n' (newline)
--------------------------------------------------------------------------------
    end                      'end'
--------------------------------------------------------------------------------
   |                        OR
--------------------------------------------------------------------------------
    \z                       the end of the string
--------------------------------------------------------------------------------
  )                        end of look-ahead

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