正则表达式来提取分隔符内的字符串

Question

我想提取分隔符（在这种情况下是括号内）的字符串发生但不是在引号（单或双）内的字符串发生。以下是我已经尝试 - 这个表达式取括号内所有出现的，这也用引号（我不想引号内的）

public class RegexMain { 
    static final String PATTERN = "\\(([^)]+)\\)"; 
    static final Pattern CONTENT = Pattern.compile(PATTERN); 
    /** 
    * @param args 
    */ 
    public static void main(String[] args) { 
     String testString = "Rhyme (Jack) and (Jill) went up the hill on \"(Peter's)\" request."; 
     Matcher match = CONTENT.matcher(testString); 
     while(match.find()) { 
      System.out.println(match.group()); // prints Jack, Jill and Peter's 
     } 
    } 
} 

Answer 1

你可以尝试

public class RegexMain { 
    static final String PATTERN = "\\(([^)]+)\\)|\"[^\"]*\""; 
    static final Pattern CONTENT = Pattern.compile(PATTERN); 
    /** 
    * @param args 
    */ 
    public static void main(String[] args) { 
     String testString = "Rhyme (Jack) and (Jill) went up the hill on \"(Peter's)\" request."; 
     Matcher match = CONTENT.matcher(testString); 
     while(match.find()) { 
      if(match.group(1) != null) { 
       System.out.println(match.group(1)); // prints Jack, Jill 
      } 
     } 
    } 
} 

此模式将匹配带引号的字符串以及括号的人，但只有那些括号内将装上去group(1)。由于+和*在正则表达式中是贪婪的，所以它宁愿匹配"(Peter's)"而不是(Peter's)。

Answer 2

注意的：这不是最终响应因为我不熟悉JAVA，但我相信它仍然可以转换成JAVA语言。

就我而言，最简单的方法是用空字符串替换字符串中的引用部分，然后查找匹配。希望你对PHP有点熟悉，这里有个想法。

$str = "Rhyme (Jack) and (Jill) went up the hill on \" (Peter's)\" request."; 

preg_match_all(
    $pat = '~(?<=\().*?(?=\))~', 
    // anything inside parentheses 
    preg_replace('~([\'"]).*?\1~','',$str), 
    // this replaces quoted strings with '' 
    $matches 
    // and assigns the result into this variable 
); 
print_r($matches[0]); 
// $matches[0] returns the matches in preg_match_all 

// [0] => Jack 
// [1] => Jill 

Answer 3

这种情况下，您可以优雅地使用look-behind和look-ahead操作符来实现您想要的功能。这里有一个Python解决方案（我总是用它在命令行上快速尝试），但正则表达式在Java代码中应该是相同的。

此正则表达式匹配前面有左括号的内容，使用正面后视，并且使用正面预见后面的右括号成功。但是，当左括号前面有一个使用负面后视的单引号或双引号时，以及当使用负面先行视图的右单引号或双引号使右闭合圆括号成功时，它会避免这些匹配。

In [1]: import re 

In [2]: s = "Rhyme (Jack) and (Jill) went up the hill on \"(Peter's)\" request." 

In [3]: re.findall(r""" 
    ...:  (?<=    # start of positive look-behind 
    ...:   (?<!   # start of negative look-behind 
    ...:    [\"\']  # avoids matching opening parenthesis preceded by single or double quote 
    ...:  )    # end of negative look-behind 
    ...:   \(   # matches opening parenthesis 
    ...: )     # end of positive look-behind 
    ...:  \w+ (?: \'\w*)? # matches whatever your content looks like (configure this yourself)    
    ...:  (?=    # start of positive look-ahead 
    ...:   \)    # matches closing parenthesis 
    ...:   (?!   # start of negative look-ahead 
    ...:    [\"\']  # avoids matching closing parenthesis succeeded by single or double quote 
    ...:  )    # end of negative look-ahead 
    ...: )     # end of positive look-ahead 
    ...:  """, 
    ...:  s, 
    ...:  flags=re.X) 
Out[3]: ['Jack', 'Jill']