从正则表达式中转义转义的量词

Question

Having trouble removing quantifiers and meta characters from a regex with sed. To complicate matters the code is intended to be used within a eval.

$word_check = "about\s*[\w\s]+";
$word_check =~ s/\\\[.+\\\]//g;
$$word_count[$#$word_count + 1] += () = $word_check =~ /(\w+)\s*/g;

Word_check is a copy of a regex. Want to reduce the regex to only the words it is to match. Using a regex to do so. Then count the words and save the result in a array. The reason behind the choice is the regex is apart of several and when working with a match ($&) the words are removed. From a eval. The code below is intended as a string.

for ( my \$i = 0; \$i < \$\$ref_word_count[\$R]; \$i++ ) {
    \$temp_str =~ s/^\\w+\\s*//;
}

OK Found the answer. Just needed to break up the sed statement some and see the square brackets as a character alone.

$word_check =~ s/i$//;
$word_check =~ s/[\.\+]//g;
$word_check =~ s/\\[sdwSWD]//g;
$word_check =~ s/[[]//g;
$word_check =~ s/[]]//g;
$word_check =~ s/[*]//g;
$word_check =~ s/[?]//g;

Answer 1

Probably this helps

\\Q quote (disable) pattern metacharacters till \\E

\\E end either case modification or quoted section, think vi

See perldoc perlre