Re: I need regex last instance help

>> On Oct 25, 2025, at 05:19, ToddAndMargo via perl6-users <perl6- 
>> [email protected]> wrote:
>>
>> Hi All,
>>
>> What am I doing wrong here:
>>
>> [2] > my Str $x = "1.40.564"; $x~~s:i/ '.' $$ .* //; print "$x\n";
>> 1.40.564
>>
>> I am after "1.40"
>>
>> This is my work around, but I'd really like to know
>> how to do it right:
>>
>> my Str $x = "1.40.564"; $x~~m:i/ ( .*? '.') (.*? ) '.' /; print "$0$1\n";
>> 1.40
>>
>> Your in confusion,
>> -T
> 

On 10/25/25 8:06 AM, William Michels wrote:
 > [0] > my $x = "1.40.564"; $x = $x ~~ m/ ^  <-[.]>+ "." <-[.]>+ /; say 
~$x;
 > 1.40
 > [0] > my $x = "1.40.564"; $x = $x.match(/ ^  <-[.]>+ "." <-[.]>+ /); say
 > ~$x;
 > 1.40
 >
 > #OR
 >
 > [1] > my $x = "1.40.564"; $x = S/ "." <-[.]>+ $ // given $x; say $x;
 > 1.40
 > [1] > my $x = "1.40.564"; $x = $x.subst(/ "." <-[.]>+ $  /, "" ); say $x;
 > 1.40
 >
 > One issue with your code was failing to create a `<-[.]>` enumerated
 > character class.
 > This (negated) `<-[.]>` class consists of all characters other than `.`
 > dot (i.e. period).
 >
 > Now read the regexes above:
 >      1. The first two matching expressions look for "start-of-string,
 > non-dot characters, a dot, then more non-dot characters".
 >      2. The last two substitution expressions look for "a dot, non-dot
 > characters, then end-of-string".
 >
 > HTH, Bill.
 >

Hi Bill,

Thank you.  It does!

It also tells me that the easiest way way actually the "m'
work around I came up with and that I have no understanding
of $$ and ^^ in regexes.  So I have some work to do.

-T