Replace regex with captured group ONLY - regex

Question

I'm trying to understand why the following does not give me what I think (or want :)) should be returned:
sed -r 's/^(.*?)(Some text)?(.*)$/\2/' list_of_values
or Perl:
perl -lpe 's/^(.*?)(Some text)?(.*)$/$2/' list_of_values
So I want my result to be just the Some text, otherwise (meaning if there was nothing captured in $2) then it should just be EMPTY.
I did notice that with perl it does work if Some text is at the start of the line/string (which baffles me...). (Also noticed that removing ^ and $ has no effect)
Basically, I'm trying to get what grep would return with the --only-matching option as discussed here. Only I want/need to use sub/replace in the regex.
EDITED (added sample data)
Sample input:
$ cat -n list_of_values
1 Black
2 Blue
3 Brown
4 Dial Color
5 Fabric
6 Leather and Some text after that ....
7 Pearl Color
8 Stainless Steel
9 White
10 White Mother-of-Pearl Some text stuff
Desired output:
$ perl -ple '$_ = /(Some text)/ ? $1 : ""' list_of_values | cat -n
1
2
3
4
5
6 Some text
7
8
9
10 Some text

Answer 1

First of all, this shows how to duplicate grep -o using Perl.
You're asking why
foo Some text bar
012345678901234567
results in just a empty string instead of
Some text
Well,
At position 0, ^ matches 0 characters.
At position 0, (.*?) matches 0 characters.
At position 0, (Some text)? matches 0 characters.
At position 0, (.*) matches 17 characters.
At position 17, $ matches 0 characters.
Match succeeds.
You could use
s{^ .*? (?: (Some[ ]text) .* | $ )}{ $1 // "" }exs;
or
s{^ .*? (?: (Some[ ]text) .* | $ )}{$1}xs; # Warns if warnings are on.
Far simpler:
$_ = /(Some text)/ ? $1 : "";
I question your use of -p. Are you sure you want a line of output for each line of input? It seems to me you'd rather have
perl -nle'print $1 if /(Some text)/'