Perl:如何提取括号之间的字符串

Question

我有一个 moinmoin 文本格式的文件:

* [[  Virtualbox Guest Additions]] (2011/10/17 15:19)
* [[  Abiword Wordprocessor]] (2010/10/27 20:17)
* [[  Sylpheed E-Mail]] (2010/03/30 21:49)
* [[   Kupfer]] (2010/05/16 20:18)

“[[”和“]]”之间的所有单词都是条目的简短描述。我需要提取整个条目，而不是每个单词。

我在这里找到了类似问题的答案:https://stackoverflow.com/a/2700749/819596但无法理解答案:"my @array = $str =~/(\{ (?: [^{}]* | (?0) )*\} )/xg;"

任何有效的东西都会被接受，但解释会有很大帮助，即:(?0) 或 /xg 的作用。

Answer 1

代码大概是这样的:

use warnings; 
use strict;

my @subjects; # declaring a lexical variable to store all the subjects
my $pattern = qr/ 
  \[ \[    # matching two `[` signs
  \s*      # ... and, if any, whitespace after them
  ([^]]+) # starting from the first non-whitespace symbol, capture all the non-']' symbols
  ]]
/x;

# main processing loop:
while (<DATA>) { # reading the source file line by line
  if (/$pattern/) {      # if line is matched by our pattern
    push @subjects, $1;  # ... push the captured group of symbols into our array
  }
}
print $_, "\n" for @subjects; # print our array of subject line by line

__DATA__
* [[  Virtualbox Guest Additions]] (2011/10/17 15:19)
* [[  Abiword Wordprocessor]] (2010/10/27 20:17)
* [[  Sylpheed E-Mail]] (2010/03/30 21:49)
* [[   Kupfer]] (2010/05/16 20:18)

如我所见，你需要的可以描述如下:在文件的每一行中尝试找到这个符号序列......

[[, an opening delimiter, 
then 0 or more whitespace symbols,
then all the symbols that make a subject (which should be saved),
then ]], a closing delimiter

如您所见，此描述很自然地转换为正则表达式。唯一可能不需要的是 /x 正则表达式修饰符，它允许我对它进行广泛的注释。 )