Regex: Interpret groups with the same content as a single group - regex

Question

I have the following situation:
^ID[ \t]*=[ \t]*('(.*)'|"(.*)")
The group with content
01
when a file contains:
ID = '01'
is the second.
Instead if:
ID = "01"
is the third.
This cause me a problem with perl:
perl -lne "print \$2 if /^ID[ \t]*=[ \t]*('(.*)'|\"(.*)\")/" test.txt
That if group with single quotes matches then i get the output:
01
Otherwise i obtain an empty string.
How do I make both the case of single quotes and double quotes interpret as group two in regex?

Answer 1

You can print both the groups, as they can never match at the same time:
perl -lne "print \$2.\$3 if /^ID[ \t]*=[ \t]*('(.*)'|\"(.*)\")/"
or remember the quotes in $2 and use $3 for the quoted string, followed by the remembered quote:
perl -lne "print \$3 if /^ID[ \t]*=[ \t]*((['\"])(.*)\2)/"

Answer 2

This looks like it's a good candidate for the branch reset operator, (?|...). Either capture in that alternation is $1, and the branch-reset construct takes care of the grouping without capturing anything:
use v5.10;
my #strings = qw( ID='01' ID="01" ID="01');
foreach ( #strings ) {
say $1 if m/^ID \h* = \h* (?|'(\d+)'|"(\d+)") /x
}
You need v5.10, and that allows you to use the \h to match horizontal whitespace.
But, you don't need to repeat the pattern. You can match the quote and match that same quote later. A relative backreference, \g{N}, can do that:
use v5.10;
my #strings = qw( ID='01' ID="01" ID="01' );
foreach ( #strings ) {
say $2 if m/^ID \h* = \h* (['"])(\d+)\g{-2} /x
}
I prefer that \g{-2} because I usually don't have to update numbering if I change the pattern to include more captures before the thing if refers to.
And, since this is a one-liner, don't type out the literal quotes (as ikegami has already shown):
say $2 if m/^ID \h* = \h* ([\x22\x27])(\d+)\g{-2} /x

Answer 3

Only one of the two will be defined, so simply use the one that's defined.
perl -nle'print $1//$2 if /^ID\h*=\h*(?:\x27(.*)\x27|"(.*)")/' # \x27 is '
You could also use a backreference.
perl -nle'print $2 if /^ID\h*=\h*(["\x27])(.*)\1/'
Note that all the provided solutions including these two fail (leave the escape sequence in) if you have something like ID="abc\"def" or ID="abc\ndef", assuming those are supported.

Answer 4

Thank you #brian_d_foy:
perl -lne "print \$1 if /^ID\h*=\h*(?|'(.*)'|\"(.*)\")/" test.txt
Or better:
perl -lne "print \$2 if /^ID\h*=\h*(['\"])(.*)\1/" test.txt
I have decided of accept also
ID = 01 #Followed by one or more horizontal spaces.
In addition to:
ID = "01" #Followed by one or more horizontal spaces.
And:
ID = '01' #Followed by one or more horizontal spaces.
Therefore I have adopted a super very complex solution:
perl -lne "print \$2 if /^ID\h*=\h*(?|(['\"])(.*)\1|(([^\h'\"]*)))\h*(?:#.*)?$/" test.txt
I have done a fusion of your both solutions #brian_d_foy. The double round parentheses are used to bring the second alternative to the second group as well, otherwise it would be the first group and without even the "branch reset operator", it would be group 4.
I after have enhanced the sintax in a function
function parse-config {
command perl -pe "s/\R/\n/g" "$2" | command perl -lne "print \$2 if /^$1\h*=\h*(?|(['\"])(.*)\1|(([^\h'\"]*)))\h*(?:#.*)?$/"
return $?
}
parse-config "ID" "test.txt"
In this:
"s/\R/\n/g"
I replace all CRLF or CR or LF, in LF. \R is a super powerfull special character present from perl v5.10. Apparently this version of perl has introduced several fundamental innovations for me. The chance would have that I needed all (\h \R ?|). Whoever did the update was brilliant.
I needed this because the dollar "$" at the end of the line did not work, because there was a "\r" before the "Linux end of line" "\n".