Perl包含续行并且忽略双引号

时间:2018-08-07 17:53:38

标签: perl double quotes lines continuation

我正在编写一个脚本,该脚本的前两行需要生成 foo ,后三行需要生成 bar 。我在这里遇到两个问题。

  1. 如何让Perl忽略第一个foo周围的双引号?

  2. 如何将反斜杠识别为续行? -

样本输入:

reset -name "foo"
quasi_static -name foo
reset \
-name bar
set_case_analysis -name "bar"

我的代码:

   if (/^\s*set_case_analysis.*\-name\s+(\S+)/) 
   {
      $set_case_analysis{$1}=1;
      print "Set Case $1\n";
   } 
   elsif (/^\s*quasi_static.*\-name\s+(\S+)/) 
   {
      $quasi_static{$1}=1;
      print "Quasi Static $1\n";
   } 
   elsif (/^\s*reset\s+.*\-name\s+(\S+)/) 
   {
      $reset{$1}=1;
      print "Reset $1\n";
    }

1 个答案:

答案 0 :(得分:0)

如果要逐行浏览文件,则可以将部分行保留在变量中,然后将它们与下一行连接。通读代码-我已经评论了该功能。

my $curr;
my $txt;
open ( IN, "<", 'yourinputfile.txt' ) or die 'Could not open file: ' . $!;
while (<IN>) {
  chomp;
  # if the line ends with a backslash, save the segment in $curr and go on to the next line
  if ( m!(.*?) \$! ) {
    $curr = $1;
    next;
  }
  # if $curr exists, add this line on to it
  if ( $curr ) {
    $curr .= $_;
  }
  # otherwise, set $curr to the line contents
  else {
    $curr = $_;
  }

  if ( $curr =~ /set_case_analysis -name\s+\"?(\S+)/) {
      # if the string is in quotes, the regex will leave the final " on the string
      # remove it
      ( $txt = $1 ) =~ s/"$//;
      print "Set Case $txt\n";
      $set_case_analysis{$txt}=1;
   }
   elsif ($curr =~ /quasi_static -name\s+(\S+)/) {
      ( $txt = $1 ) =~ s/"$//;
      print "Quasi Static $txt\n";
      $quasi_static{$txt}=1;

   }
   elsif ($curr =~ /reset .*?-name\s+\"?(\S+)/) {
      ( $txt = $1 ) =~ s/"$//;
      print "Reset $txt\n";
      $reset{$txt}=1;
  }
  # reset $curr
  $curr = '';
}

您可以通过执行以下操作使其更紧凑,更整洁:

if ( $curr =~ /(\w+) -name \"?(\S+)/) {
      ( $txt = $2 ) =~ s/"$//;
      $data{$1}{$txt}=1;
}

您将获得带有三个键set_case_analysisquasi_staticreset以及来自-name的各种不同值的嵌套哈希结构。

%data = (
  quasi_static => ( foo => 1, bar => 1 ),
  reset => ( pip => 1, pap => 1, pop => 1 ),
  set_case_analysis => ( foo => 1, bar => 1 )
);