在Perl中，如何处理配置文件中的续行？

Question

So I'm trying to read in a config. file in Perl. The config file uses a trailing backslash to indicate a line continuation. For instance, the file might look like this:

  === somefile ===
  foo=bar
  x=this\
  is\
  a\
  multiline statement.

I have code that reads in the file, and then processes the trailing backslash(es) to concatenate the lines. However, it looks like Perl already did it for me. For instance, the code:

  open(fh, 'somefile');
  @data = <fh>;
  print join('', @data);

prints:

  foo=bar
  x=thisisamultiline statement

Lo and behold, the '@data = ;' statement appears to have already handled the trailing backslash!

Is this defined behavior in Perl?

Answer 1

I have no idea what you are seeing, but that is not valid Perl code and that is not a behavior in Perl. Here is some Perl code that does what you want:

#!/usr/bin/perl

use strict;
use warnings;

while (my $line = <DATA>) {
    #collapse lines that end with \
    while ($line =~ s/\\\n//) {
        $line .= <DATA>;
    }
    print $line;
}

__DATA__
foo=bar
x=this\
is\
a\
multiline statement.

Note: If you are typing the file in on the commandline like this:

perl -ple 1 <<!
foo\
bar
baz
!

Then you are seeing the effect of your shell, not Perl. Consider the following counterexample:

printf 'foo\\\nbar\nbaz\n' | perl -ple 1

Answer 2

My ConfigReader::Simple module supports continuation lines in config files, and should handle your config if it's the format in your question.

If you want to see how to do it yourself, check out the source for that module. It's not a lot of code.

Answer 3

I don't know what exactly you are doing, but the code you gave us doesn't even run:

=> cat z.pl
#!/usr/bin/perl
fh = open('somefile', 'r');
@data = <fh>;
print join('', @data);

=> perl z.pl
Can't modify constant item in scalar assignment at z.pl line 2, near ");"
Execution of z.pl aborted due to compilation errors.

And if I change the snippet to be actual perl:

=> cat z.pl
#!/usr/bin/perl
open my $fh, '<', 'somefile';
my @data = <$fh>;
print join('', @data);

it clearly doesn't mangle the data:

=> perl z.pl
foo=bar
x=this\
is\
a\
multiline statement.