This will not work (the . will match the /'s and preface them with 0's)... try \d instead:
$ perl -e '$a="9/9/1973\n";$a=~s/\b(.)\b/0$1/g;print $a;' 090/090/1973 $ perl -e '$a="9/9/1973\n";$a=~s/\b(\d)\b/0$1/g;print $a;' 09/09/1973 $ perl -e '$a="9/09/1973\n";$a=~s/\b(\d)\b/0$1/g;print $a;' 09/09/1973 I would use printf's, tho, even tho I love regexps. :) Jason ----- Original Message ----- From: "Paul Makepeace" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Thursday, January 31, 2002 8:47 PM Subject: Re: substitution question > On Thu, Jan 31, 2002 at 05:41:15PM -0800, Pradeep Sethi wrote: > > Thanks but I am looking of any regexp substitution. > > > > sorry for typo : I need to change 9/9/1973 to 09/09/1973 > > How about, > > s/\b(.)\b/0$1/g > > Paul >
