You're probably better off using good ole' substr instead. It's built for doing such things.

If you really want to use a regex for some perverse reason -- well, then you're stark raving mad. But in good company. Anyways, think on the following:

Putting that all together, we would get $callerid =~ /(.{7})$/; which would leave the last seven digits in $1. I will reiterate that you don't want to do this, though. It's slow and ugly; use substr. The above was provided as a learning exercise.

Share and enjoy. :)

perl -pe '"I lo*`+$^X$\"$]!$/"=~m%(.*)%s;$_=$1;y^`+*^e v^#$&V"+@( NO CARRIER'

You can code exactly like that in Perl, thanks to the /x modifier. It allows you to insert comments inside the regexp.

$callerid =~ / # we want to match
             ( # and remember
             . # any character
           {7} # 7 times
             ) # stop remembering
             $ # then the end of line
             /x; # and that's the end of the match

In this particular case, this is overkill, but in some cases, it can be very useful in more cmoplicated situations.

And as all others in this thread have said, I too recommend using substr.

All you need is substr

$callerid = '12345678901234567';
$last_bit = substr $callerid, -7;
print $last_bit;
[download]

cheers

tachyon

s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

Use substr: $se7en = substr $callerid, -7.

If you really insist, you could say ($se7en) = ($callerid =~ /(.{7})$/);. But you really want to use a regexp only if you want to validate $callerid at the same time (for instance, to ensure the last 7 characters really are digits). To do that, you'll have to read perlre.