#!/usr/bin/perl $num=14.45905495; @post=$1 if ($num=~/(\d+\.\d\d)\d+/); print "$post[0]\n";
output ---- 14.45 explanation ------ \d+ = at least one digit or more \. = followed by a period. \d\d = two digits \d+ = followed by any number of digits (\d+\.\d\d) Sets the portion in parens to be memory location $1. Hope this helps. I am sure there is better way but this should do it. Paul > -----Original Message----- > From: David O'Dell [mailto:[EMAIL PROTECTED]] > Sent: Tuesday, December 31, 2002 1:16 PM > To: [EMAIL PROTECTED] > Subject: substitution or matching regex > > > I have a variable: > $NUM = '14.45905495'; > and I want to remove the trailing digits and only leave 2 > after the period so it ends up > > '14.45' > > I've tried to do this but it appears to return as an array > and always prints out "1". > > #!perl -w > > $NUM = '14.45905495'; > @POST = ($NUM =~ /\d\.\d{2}/); > print "new value is $POST[0]\n"; > > after being run > > > $ ./test_ex.pl > new value is 1 > > > > > -- > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
