perl: how to insert numerical digit immediately after regexp backreference variable?

chadwick · 05-02-2008, 03:52 PM

I'm trying to insert the digit 0 (zero) into a string, as the following.
Convert:

xxxxxxx-d-xxxxxx.mp3

to

xxxxxxx-0d-xxxxxx.mp3

where d is a decimal digit and x means I don't care what it is.

If I do it like:

s/(.*-)(\d)(-.*mp3)/$1 0$2$3/;

then it prints a space where I have a space.

If I do it like:

s/(.*-)(\d)(-.*mp3)/$10$2$3/;

then it doesn't end up using $1, possibly since it thinks I'm trying to use $10 which doesn't exist.

If I do it like:

s/(.*-)(\d)(-.*mp3)/{$1}0$2$3/;
s/(.*-)(\d)(-.*mp3)/($1)0$2$3/;
s/(.*-)(\d)(-.*mp3)/$1{0}$2$3/;
s/(.*-)(\d)(-.*mp3)/$1(0)$2$3/;

Then it prints the brackets which I don't want.
I've tried using named backreferences instead but I can't get that to work. Not sure if I'm using the wrong syntax or the wrong version of perl. I'm using perl v5.8.8

Thanks

forrestt · 05-02-2008, 04:03 PM

Two ways to do it (I'm sure there are others, and probably better):

Code:

$zero='0';
$file =~ s/(.*-)(\d)(-.*mp3)/$1$zero$2$3/;

or

Code:

$file =~ s/(.*-)(\d)(-.*mp3)/$1 $2$3/;
$file =~ s/ /0/;

HTH

Forrest

nifund9 · 05-02-2008, 04:05 PM

My Perl is a little rusty, but is it:

s/(.*-)(\d)(-.*mp3)/$1\0$2$3/;

forrestt · 05-02-2008, 04:17 PM

One more if you want to have it all on one line:
the "-" is just to have a character, and the \b deletes it:

Code:

s/(.*-)(\d)(-.*mp3)/$1-\b0$2$3/;

HTH

Forrest

chadwick · 05-02-2008, 04:19 PM

Thanks. I went with the $zero='0' method which worked really well.
I tried the \0 method as well but that didn't seem to do it.

chadwick · 05-02-2008, 04:23 PM

Quote:

Originally Posted by forrestt

the "-" is just to have a character, and the \b deletes it:

Nice, I'd never seen that before. I'll remember it for next time.

bartonski · 05-18-2008, 10:13 PM

Quote:

Originally Posted by nifund9

My Perl is a little rusty, but is it:

s/(.*-)(\d)(-.*mp3)/$1\0$2$3/;

One more way to do it, this being Perl, after all:

Code:

 s/(.*-)(\d)(-.*mp3)/${1}0$2$3/;

I haven't actually tested this, but it's the way that you would handle a variable outside of a regular expression, eg.

Code:

$foo = "ASDF";
$foobar = "You Lose";
print "${foo}bar";  # prints "ASDFbar"
print "$foobar";    # prints "You Lose"

chadwick · 05-18-2008, 11:18 PM

Quote:

Originally Posted by bartonski

One more way to do it, this being Perl, after all:

Code:

 s/(.*-)(\d)(-.*mp3)/${1}0$2$3/;

I haven't actually tested this, but it's the way that you would handle a variable outside of a regular expression, eg. ...

Yes this makes sense. It's what I was trying to do with my brackets above but couldn't remember how to do it properly and ended up thinking that I was making stuff up. Thanks for pointing that out.

osor · 05-19-2008, 12:49 PM

Btw, you can avoid the issue with 0-width assertions (which are an oft-overlooked part of Perl’s regular expressions). For example,

Code:

s/(?<=-)(?=\d-.*mp3)/0/