Perl 包含连续行并忽略双引号-解网

问：

我正在编写一个脚本，需要为前两行生成 foo，为后三行生成 bar。我在这里遇到了两个问题。

如何让 Perl 忽略第一个 foo 周围的双引号？
如何让它将反斜杠识别为延续行？-

输入示例：

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";
    }

Perl 双引号行延续

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 = '';
}

你可以通过做这样的事情来使它更加紧凑和整洁：

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

您将获得一个嵌套的哈希结构，其中包含三个键、、和，以及中的各种不同值。set_case_analysisquasi_staticreset-name

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

Perl 包含连续行并忽略双引号

Perl include continuation lines and ignore double quotes

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