while(<>) {
    s/(^1.*)/\1,P/g;
    print;
}
[download]

In Perl, we use $1 instead of \1 in the substitution string. The right hand side is just another interpolated string.

From perlop:


It is at this step that "\1" is begrudgingly converted to $1 in the replacement text of "s///" to correct the incorrigible sed hackers who haven't picked up the saner idiom yet. A warning is emitted if the "use warnings" pragma or the -w command-line flag (that is, the $^W variable) was set.

From Jeffrey E. F. Friedl's great book Mastering Regular Expressions:


The Perl manpage makes a concerted effort to point out that \1 is not available as a backreerence outside of a regex. (Use the variable $1 instead.) \1 is much more than a simple notational convenience - the variable $1 refers to a string of static text matched during some previously completed successful match. On the other hand, \1 is a true regex metacharacter to match text similar to that matched within the first parenthesized subexpression (...)


perl -we's//\1/' 2>&1 |splain

\1 better written as $1 at -e line 1 (#1)
(W syntax) Outside of patterns, backreferences live on as variables. The use of backslashes is grandfathered on the right-hand side of a substitution, but stylistically it's better to use the variable form because other Perl programmers will expect it, and it works better if there are more than 9 backreferences.

And dot star (.*) is bad in this case.

2;0 juerd@ouranos:~$ perl -e'undef christmas'
Segmentation fault
2;139 juerd@ouranos:~$
[download]

Thanks!

I knew I'd learn something from Perlmonks!

--
Its like a dog that can sing and dance.
It's remarkable because it can do it.
Not that it can do it well.

Would this work:

s/^1/^$,P/g
[download]

 

Can you calrify yours please. Thanks..

my $text = "1..., some words, Dec 13, 2001";


if ($text =~ m/^\1/) {
$text =~ m/(.*$)/;
print "${1},P";
}
[download]

Thanks ;-}

Nooooo! :) See Death to dot star to read why .* is bad.
You can't use \1 there, you should just use 1.
Please note that /^1/ will also match strings beginning with 123.

I suggest the following solution:

$line =~ s/$/,P/ if /^1\D/;
[download]



Which will substitute the end of the line (which is _before_ a newline if there is any) with ,P (and thus "append" it) if the line starts with 1 followed by anything-that-is-not-a-digit.

2;0 juerd@ouranos:~$ perl -e'undef christmas'
Segmentation fault
2;139 juerd@ouranos:~$
[download]

Or, even shorter (and hopefully clearer) ..

my $text = "1..., some words, Dec 13, 2001";
print "$text,P" if ( $text =~ /^1/ );
[download]

Although you may want \n at the end of the line.

--t. alex

"Excellent. Release the hounds." -- Monty Burns.