RE: substitution or matching regex

2002-12-31 Thread wiggins
You may want to do this with sprintf rather than a regex, check out: perldoc -f sprintf I believe because your regex isn't grouping any terms it is returning a status of true or false, aka 1 for true since your regex does in fact match. You might try paren's around the whole thing if you still

RE: substitution or matching regex

2002-12-31 Thread Kipp, James
> 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'; > @POS

RE: substitution or matching regex

2002-12-31 Thread Paul Kraus
#!/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 memo