Question

我正在尝试匹配perl中的模式并需要一些帮助。

我需要从字符串中删除任何与[xxxx]相匹配的内容，即在其中打开括号内的东西 - 首先出现的右括号。

所以我试图用空格替换开口括号，里面的东西，首先用以下代码关闭括号：

   if($_ =~ /[/)
  {
    print "In here!\n";
    $_ =~ s/[(.*?)]/ /ig;
  }

同样地，我需要匹配它内部的角括号 - 首先关闭角括号。

我正在使用以下代码执行此操作：

   if($_ =~ /</)
  {
    print "In here!\n";
    $_ =~ s/<(.*?)>/ /ig;
  }

这一些似乎没有用。我的样本数据如下：

 'Joanne' <!--Her name does NOT contain "Kathleen"; see the section "Name"--> "'Jo'" 'Rowling', OBE [http://news bbc co uk/1/hi/uk/793844 stm Caine heads birthday honours list]  BBC News  17 June 2000  Retrieved 25 October 2000  , [http://content scholastic com/browse/contributor jsp?id=3578 JK Rowling Biography]  Scholastic com  Retrieved 20 October 2007  better known as 'J  K  Rowling' ,<ref name=telegraph>[http://www telegraph co uk/news/uknews/1531779/BBCs-secret-guide-to-avoid-tripping-over-your-tongue html Daily Telegraph, BBC's secret guide to avoid tripping over your tongue, 19 October 2006] is a British <!--do not change to "English" or "Scottish" until issue is resolved --> author best known as the creator of the [[Harry Potter]] fantasy series, the idea for which was conceived whilst on a train trip from Manchester to London in 1990  The Potter books have gained worldwide attention, won multiple awards, sold more than 400 million copies and been the basis for a popular series of films, in which Rowling had creative control serving as a producer in two of the seven installments  [http://www businesswire com/news/home/20100920005538/en/Warner-Bros -Pictures-Worldwide-Satellite-Trailer-Debut%C2%A0Harry Business Wire - Warner Bros  Pictures mentions J  K  Rowling as producer ]

任何帮助将不胜感激。谢谢！

Answer 1

你需要使用它：

1 while s/\[[^\[\]]*\];

演示：

% echo "i have [some [square] brackets] in [here] and [here] today."| perl -pe '1 while s/\[[^\[\]]*\]/NADA/g'
i have NADA in NADA and NADA today.

与失败的对比：

% echo "i have [some [square] brackets] in [here] and [here] today." | perl -pe 's/\[.*?\]/NADA/g'
i have NADA brackets] in NADA and NADA today.

我留下的递归正则表达式作为读者的练习。：）

编辑： Eric Strom提供了一个递归解决方案，您不必使用1 while：

% echo "i have [some [square] brackets] in [here] and [here] today." | perl -pe 's/\[(?:[^\[\]]*|(?R))*\]/NADA/g'
i have NADA in NADA and NADA today.

Answer 2

方括号在正则表达式语法中具有特殊含义，因此请转义它们：/\[.*?\]/。（你也不需要这里的括号，做不区分大小写的匹配是没有意义的。）
自从我不得不与Perl搏斗已经很长时间了，但我很确定用正则表达式测试$ _也会修改$ _（即使你没有使用s ///）。你无论如何都不需要测试;只需运行替换，如果模式在任何地方都不匹配，那么它就不会做任何事情。

Answer 3

$_ =~ /someregex/不会修改$_

只需注意，$_ =~ /someregex/和/someregex/也会做同样的事情。

此外，您不需要检查是否存在[或＆lt;或分组括号：

s/\[.*?\]/ /g;

s/<.*?>/ /g;

将完成你想要的工作。

编辑：更改了代码以匹配您正在修改$ _

的事实

Perl模式匹配问题

3 个答案: