Question

输入文件包含多个换行符，空标记如下：

<html>
<body>
<title>XXX</title>
<p>text...</p>
<collaboration seq="">
<ce:text></ce:text>
</collaboration>
...


<p>text</p>
<collaboration seq="">
<ce:text>AAA</ce:text>
</collaboration>
<p>text</p>
</body>
</html>

只需一个换行符所需的输出文件，必须删除空标记

<html>
<body>
<title>XXX</title>
<p>text...</p>
...
<p>text</p>  
<p>text</p>
<collaboration seq="">
<ce:text>AAA</ce:text>
</collaboration>
</body>
</html>

已经尝试过的编码：

print "Enter the file name without extension: ";
chomp($filename=<STDIN>);
open(RED,"$filename.txt") || die "Could not open TXT file";
open(WRIT,">$filename.html");
while(<RED>)
{
  #process in file
  s/<collaboration seq="">\n<ce:text><\/ce:text>\n<\/collaboration>//g;
  s/\n\n//g;
  print WRIT $_;
}
close(RED);
close(WRIT);

以上编码不会清除所需的任何内容......如何解决？

Answer 1

首先你应该实际上啜饮文件。所以我们假设您使用zigdon's method：

my $file;
{
    print "Enter the file name without extension: ";
    my $filename = <STDIN>

    chomp($filename);
    open F, $filename or die "Can't read $filename: $!";
    local $/;  # enable slurp mode, locally.
    $file = <F>;
    close F;
}

现在$file包含文件的内容，因此您可以使用它。

#process in file
$file ~= s/<collaboration seq="">\R<ce:text><\/ce:text>\R<\/collaboration>//g;
$file ~= s/\R{2,}/\n/g; #I'm guessing this is probably what you intended
print WRIT $file;

Answer 2

您可以使用XML :: Simple：

# use XML simple to process the XML
my $xs = XML::Simple->new(
      # remove extra whitespace
      NormaliseSpace => 2,
      # keep root element
      KeepRoot       => 1,
      # force elements to arrays
      ForceArray     => 1,
      # ignore empty elements
      SuppressEmpty  => 1
);
# read in the XML
my $ref = $xs->XMLin($xml);


# print out the XML minus the empty tags
print $xs->XMLout($ref);

如何使用Perl在RE中替换s ///中使用换行符？

2 个答案: