perl – 如何在文本中提取所有引用？

发布时间：2020-12-15 23:35:04 所属栏目：大数据来源：网络整理

导读：我正在寻找一个输出文本中所有引用的SimpleGrepSedPerlOr PythonOneLiner. 例1： echo “HAL,” noted Frank,“said that everything was going extremely well.” | SimpleGrepSedPerlOrPythonOneLiner 标准输出： "HAL,""said that everything was going e

我正在寻找一个输出文本中所有引用的SimpleGrepSedPerlOr PythonOneLiner.

例1：

echo “HAL,” noted Frank,“said that everything was going extremely well.” | SimpleGrepSedPerlOrPythonOneLiner

标准输出：

"HAL,"
"said that everything was going extremely well.”

例2：

cat MicrosoftWindowsXPEula.txt | SimpleGrepSedPerlOrPythonOneLiner

标准输出：

"EULA"
"Software"
"Workstation Computer"
"Device"
"DRM"

等等

(link to the corresponding text).

解决方法

我喜欢这个：

perl -ne 'print "$_n" foreach /"((?>[^"]|+[^"]|(?:\)*")*)"/g;'

它有点冗长,但它比最简单的实现更好地处理转义引用和回溯.它的意思是：

my $re = qr{
   "               # Begin it with literal quote
   ( 
     (?>           # prevent backtracking once the alternation has been
                   # satisfied. It either agrees or it does not. This expression
                   # only needs one direction,or we fail out of the branch

         [^"]    # a character that is not a dquote or a backslash
     |   +       # OR if a backslash,then any number of backslashes followed by 
         [^"]      # something that is not a quote
     |           # OR again a backslash
         (?>\)* # followed by any number of *pairs* of backslashes (as units)
         "         # and a quote
     )*            # any number of *set* qualifying phrases
  )                # all batched up together
  "                # Ended by a literal quote
}x;

如果你不需要那么大的力量 – 说它只是可能是对话而不是结构化的引用,那么

/"([^"]*)"/

可能与其他任何东西一样有效.

（编辑：李大同）

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