Question

I have a string like this:

$var = '123456d';

or possibly

$var = '123456'; # (no alpha char)

There will be up to 6 digits no less than 5 if that matters.
No alpha, spaces or special mixed in the numbers.

In my Perl script, I need to remove and return the last alpha character AND the string without the alpha character.

I suppose can use \D and trim? or chop? but, there must be an easy way to get both vars quickly.

my $numbers =~   s/\D+?$var//;  #???  

my $alpha = substr($var, 0, -1); ## but no alpha check.

or

my $alpha = chop($var); ## but no alpha check.

then if a-Z and so on to check if it is an alpha character.

Or the limit of my ability solution:

$alpha = $var;
$alpha =~ s/[0-9]//ig;
$var =~ s/[a-Z]//ig;
$numbers = $var;

so result:

($numbers == '123456')
($alpha eq 'd') (or '' if nothing)

I feel like this is too trivial to ask here but, I just cannot find or write an applicable solution.

My use of =~ s is just a guess but, there must be a better way or even a one liner.

Thanks for all the help here…

SORRY after thought! There may be 2 alpha characters after the digits! They would need to be returned together as a single var ($alpha eq ‘dd’)

I found this too to0:

$var =~ s{^([0-9]+).*}{$1}i;

Answer 1

Using \pL will avoid capturing numbers, and making the string optional will assure that the match happens even for numbers without ending letters.

my $var = '123456d';
my ($num, $letter) = $var =~ /^(\d+)(\pL)?$/;

Note that you cannot use \w to capture the possible ending letter, because it also includes numbers. As tchrist has pointed out, \p{alpha} can also be used.