Sorry, that last line was wrong...
my $in = '000007032003';
$in =~ s/^0+/$1/ if (/(\d{2})(\d{2})(\d{4})$/);
# my $out = qq($in/$2/$3); yuck, sorry
# Now we can do the replacement with no problem
# because we've restructured $in to dd/dd/dddd
my $out = qq($1/$2/$3) if (/(\d{2})(\d{2})(\d{4})/);
-----Original Message-----
From: Scot Robnett [mailto:[EMAIL PROTECTED]
Sent: Wednesday, July 02, 2003 2:03 PM
To: Paul Kraus; Sara; [EMAIL PROTECTED]
Subject: RE: Regex question.
> > 2- Want to format dates like birth = 02151956 should be 02/15/1956
> my $date = "$1/$2/$3/" if (/(\d\d)(\d\d)(\d\d\d\d)/)
# All of this is UNTESTED, please treat as such.
# More of "the same but different"
my $date = qq($1/$2/$3) if /(\d{2})(\d{2})(\d{4})/;
# Takes into account dates like 07/03/2003, unless
# of course you've lopped off the preceding "000" type
# strings with s/^0+//; --- for example:
my $in = '000007032003';
my $out = qq($1/$2/$3) if /(\d{2})(\d{2})(\d{4})/;
print $out;
# would result in printing:
# 00/00/0703
# This might work better (preserving preceding "0" on the date)
my $in = '000007032003';
$in =~ s/^0+/$1/ if (/(\d{2})(\d{2})(\d{4})$/);
my $out = qq($in/$2/$3);
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