Bryan Harris wrote:
>
> Is there a way to interpolate strings that the user enters?
>
> I'm writing a filter (pipe-cat = pat) that lets you add text to the front or
> end of some piped data:
>
> echo "2" | pat "1" - "3\n"
>
> The "-" represents the piped data, so the above should print:
>
> % echo "2" | pat "1\n" - "3\n"
> 1
> 2
> 3
> %
>
> But here's what I get:
>
> % echo "2" | pat "1" - "3\n"
> 1\n2
> 3\n%
>
> How can I interpolate the "\n" that the user entered?
$string =~ s/\\n/\n/g;
> Here's my code (disclaimer: I'm still very much a novice at this stuff).
>
> #! /usr/bin/perl -w
>
> $newtxt = "";
> while ($_ = shift) {
> if ($_ eq "-") {
> open(FILE, "-") || die("Couldn't read from STDIN: $!\n");
> undef $/;
> $newtxt .= <FILE>;
> close(FILE);
Since you are reading from STDIN and it is already open:
if ( $_ eq '-' ) {
local $/;
$newtxt .= <STDIN>;
> }
> else { $newtxt .= $_; }
> }
> $/ = "\n";
If you had used 'local $/;' above then you wouldn't have to set it here.
> chomp $newtxt;
> print $newtxt, "\n";
Why remove "\n" in one line and then add it back on the next line?
> exit(0);
>
> I'm also very much open to tips from the pros on this stuff.
You could do it something like this:
#!/usr/bin/perl
use warnings;
use strict;
my %escapes = (
'\\' => "\\",
n => "\n",
t => "\t",
f => "\f",
b => "\b",
r => "\r",
);
my $newtxt = join '', map {
s/\\([\\ntfbr])/$escapes{$1}/g;
local $/;
$_ eq '-' ? <STDIN> : $_
} @ARGV;
print $newtxt;
exit 0;
__END__
John
--
use Perl;
program
fulfillment
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