Thank you very much Japhy for this solution now how can I replace digits with formatted Xs like :
123412341234 to XXXX-XXXX-1234 and 0123412341234 to XXXX-XXXX-X1234 using regular expressions. Thanks in Advance Pradeep -----Original Message----- From: Jeff 'japhy' Pinyan [mailto:[EMAIL PROTECTED]] Sent: Monday, March 04, 2002 3:55 PM To: Pradeep Sethi Cc: [EMAIL PROTECTED] Subject: Re: replacing last 4 digits On Mar 4, Sethi, Pradeep said: >I have a number 342389842452. > >how do a substitute of everything with X but last 4 digits using regular >expressions > >like xxxxxxxx2452 You could take an approach like: s/\d(?=\d{4})/x/g; The (?=...) means "look ahead for ...". So this regex matches a digit, and then checks to see if it CAN match four more digits, without really going there in the string. If it matches, it replaces the digit with an "x". A much simpler approach to understand is: s/(\d+)(\d{4})/"x" x length($1) . $2/e; This time, we match all the digits we can to $1, allowing us to back up a little to match four digits into $2. Then we replace ALL of that with length($1) x's, followed by the four digits saved in $2. The /e modifier means that the replacement is code to be executed. -- Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/ RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/ ** Look for "Regular Expressions in Perl" published by Manning, in 2002 ** <stu> what does y/// stand for? <tenderpuss> why, yansliterate of course. [ I'm looking for programming work. If you like my work, let me know. ] -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
