Perl regex to extract floating number

Question

I need to modify someone\'s Perl script, and I am not familiar with Perl at all.

There is a scalar variable $var, whose value is a floating point number pos

Answer 1

So I am trying suggested expressions, I guess I am not using them correctly in Perl:

my $var = "123.456.66";
print "old $var\n";
$var =~ s/^\d+(?:\.\d+)?//;
print "new $var\n";

Output:

$perl main.pl
old 123.456.66
new .66

Answer 2

As I understand it, you need to extract the first one or two groups of digits from a string. Like so.

123.456.789  # 123.456
123.456abc   # 123.456
123abc       # 123
abc123       # nothing

The regex would look like this, expanded out for a better explanation.

qr{
  (
    \d+ 
    (?: \.\d+ )?
  )
}x;

qr is the regex quoting operator. Using x means to ignore spaces so things are more readable.

\d matches digits. + says to match 1 or more of the preceding. So \d+ is 1 or more digits.

() captures the contents.

(?:) groups the contents but does not capture.

? says to capture 0 or 1 of the preceding. It means it's optional.

So (?: \.\d+ )? means a dot followed by some digits is optional.

You'd use it like so.

my $str = "123.456abc";
my $digits_re = qr{ (\d+ (?: \.\d+ )?) }x;
my($digits) = $str =~ $digits_re;
print $digits;

For more information see the Perl Regex Tutorial and you can play with it on Regex 101.

Answer 3

Following your update, the expression you need is:

^\d+(?:\.\d+)?

• ^\d+ Match digits at start of string.
• (?: Start of non capturing group.
• \.\d+ Match a literal ., followed by digits.
• )? Close non capturing group making it optional.

Check the expression here.

A Perl example:

$var = "123.456.789";
print "old $var\n";
$var =~ /(^\d+(?:\.\d+)?)/;
print "new $1\n";

Prints:

old 123.456.789
new 123.456