Remove leading zeroes but not all zeroes

Question

How can I remove all the leading zeroes but leave a final zero if the value only contains zeroes?

for example:

my $number = \"0000\";

Answer 1

Just add 0 to the string so that it's implicitly converted to a number:

my $var = '0000035600000';
$var += 0;

Answer 2

This should work:

$number =~ s/^0*(\d+)$/$1/;

0    -> 0
0000 -> 0
0001 -> 1

Edit: turns out that's just a bit too complicated. This should also work:

$number =~ s/0*(\d+)/$1/;

but I'm not sure which is better, depends on the use case.

Do check out the answer from Oesor: it's pretty sweet too, no regex involved.

Answer 3

If your $number is an integer, another way is to use printf:

for my $number (qw(005 05 5 000 0 500)) {
    printf "%d\n", $number;
}

__END__

5
5
5
0
0
500

Edit: As ysth points out, it works for all integers, not just positive integers as I originally stated.

Answer 4

Well if you want to replace all by one just change to s/^0+/0/ ?

However, this will replace all leading 0's by one 0 so if you have leading zeros before numbers you will need Mats construct.

Answer 5

$number =~ s/^0+(?=\d)//;

The positive lookahead ensures there's at least one digit left, and using + instead of * means the substitution is skipped if there are no leading zeroes.

If you're trying to normalize groups of numbers in a single string, just use \b instead of ^, and do a global substitution:

my $data = "0225 0000 030";
$data =~ s/\b0+(?=\d)//g;

Answer 6

Some answers are implicitly assuming there are only digits present or that $number is in fact a number. If this is not the case:

s/^0+(?=.)//s

Or even:

substr($number,0,-1) =~ s/^0+//s