Perl字符串模式匹配的负正则表达式

Question

我有这个正则表达式:

if($string =~ m/^(Clinton|[^Bush]|Reagan)/i)
  {print "$string\n"};

我想与克林顿和里根相提并论,但不是布什.

它不起作用.

Answer 1

你的正则表达式不起作用,因为[]定义了一个字符类,但你想要的是一个前瞻:

(?=) - Positive look ahead assertion foo(?=bar) matches foo when followed by bar
(?!) - Negative look ahead assertion foo(?!bar) matches foo when not followed by bar
(?<=) - Positive look behind assertion (?<=foo)bar matches bar when preceded by foo
(?<!) - Negative look behind assertion (?<!foo)bar matches bar when NOT preceded by foo
(?>) - Once-only subpatterns (?>\d+)bar Performance enhancing when bar not present
(?(x)) - Conditional subpatterns
(?(3)foo|fu)bar - Matches foo if 3rd subpattern has matched, fu if not
(?#) - Comment (?# Pattern does x y or z)

所以试试:(？!bush)

Answer 2

示范文本:

克林顿说,
布什用蜡笔
忘了里根

只是省略布什比赛:

$ perl -ne 'print if /^(Clinton|Reagan)/' textfile
Clinton said
Reagan forgot

或者,如果您真的想指定:

$ perl -ne 'print if /^(?!Bush)(Clinton|Reagan)/' textfile
Clinton said
Reagan forgot

Answer 3

你的正则表达式如下:

/^         - if the line starts with
(          - start a capture group
Clinton|   - "Clinton" 
|          - or
[^Bush]    - Any single character except "B", "u", "s" or "h"
|          - or
Reagan)   - "Reagan". End capture group.
/i         - Make matches case-insensitive 

所以,换句话说,正则表达式的中间部分正在搞砸你.由于它是一种"全能型"群体,它将允许任何不以"布什"中的任何大写或小写字母开头的行.例如,这些行符合您的正则表达式:

Our president, George Bush
In the news today, pigs can fly
012-3123 33

如前所述,您要么做出否定的预测,要么只是制作两个正则表达式:

if( ($string =~ m/^(Clinton|Reagan)/i) and
    ($string !~ m/^Bush/i) ) {
   print "$string\n";
}

正如mirod在评论中指出的那样,当使用插入符号(^)仅匹配行的开头时,第二次检查是非常不必要的,因为以"克林顿"或"里根"开头的行永远不会以"布什"开头.

但是,如果没有插入符号,它将是有效的.