[newbie] regexp question

[grikdog]

How do I write a regexp that will select 1.23 or -12.34 but not 1.2.3?

$a =~ s/(\-*\d+\.\d+)/foo\1bar/g;

This always selects 1.2 out of 1.2.3, and it should not. I.e., "foo1.2bar.3" is an error.

Thanks!

[numberwhun]

Maybe its because its after midnight and I have been coding all day, who knows, but I am not really sure what you are wanting to do.

That aside though, I took and modified your regex (only slightly), at the behest of the Komodo editor that I am using. Here is the quick script I wrote:

use strict;
use warnings;
 
print("Please input a string: ");
my $input = <STDIN>;
chomp($input);
 
$input =~ s/(\-*\d+\.\d+)/foo$1bar/g;
 
print("$input");
 

Now, I ran the script a few times, using the input you provided. You can see here what the script produced:

######## Script Output ############

C:\>perl select.pl

Please input a string: foo1.2bar.3
foofoo1.2barbar.3

C:\>perl select.pl

Please input a string: 1.2.3
foo1.2bar.3

C:\>perl select.pl

Please input a string: 1.23
foo1.23bar

C:\>perl select.pl

Please input a string: -12.34
foo-12.34bar

C:\>

######## End Output #############

I addeed blank lines here manually for readability of the output. As you can see, I think it did what you wanted, but again, I wasn't quite sure what you were looking for.

If this isn't what you wanted, maybe you could elaborate a fair amount and provide better examples of what you are trying to achieve.

Regards,

Jeff

[miller]

I also do not know what you're ultimately trying to acheive. However, the following faq link will probably be helpful in some way.

perldoc perlfaq4 How do I determine whether a scalar is a number/whole/integer/float?

- Miller

[KevinADC]

you never told the regexp where to stop substituting, you only gave it part of a pattern to find and replace. If you don't want the rest of the patter/string included on the replacement side you have to add it to the search pattern and exclude it in the replacement pattern.

$a = '1.2.3';
$a =~ s/(\-?\d+\.\d+)(?:\.\d)/foo\1bar/;
print $a;

adding (?:\.\d) after the search pattern and not carrying it over to the replacement pattern (using $2 or \2) prevents the regexp from including that part of the string in the replacement pattern.