How can regex ignore escaped-quotes when matching strings?

escaping, negative-lookbehind, regex

Solution

<?php
$backslash = '\\';

$pattern = <<< PATTERN
#(["'])(?:{$backslash}{$backslash}?+.)*?{$backslash}1#
PATTERN;

foreach(array(
    "<?php \$s = 'Hi everyone, we\\'re ready now.'; ?>",
    '<?php $s = "Hi everyone, we\\"re ready now."; ?>',
    "xyz'a\\'bc\\d'123",
    "x = 'My string ends with with a backslash\\\\';"
    ) as $subject) {
        preg_match($pattern, $subject, $matches);
        echo $subject , ' => ', $matches[0], "\n\n";
}

prints

<?php $s = 'Hi everyone, we\'re ready now.'; ?> => 'Hi everyone, we\'re ready now.'

<?php $s = "Hi everyone, we\"re ready now."; ?> => "Hi everyone, we\"re ready now."

xyz'a\'bc\d'123 => 'a\'bc\d'

x = 'My string ends with with a backslash\\'; => 'My string ends with with a backslash\\'

Problem

I'm trying to write a regex that will match everything BUT an apostrophe that has not been escaped. Consider the following: ``` <?php $s = 'Hi everyone, we\'re ready now.'; ?> ``` My goal is to write a regular expression that will essentially match the string portion of that. I'm thinking of something such as ``` /.*'([^']).*/ ``` in order to match a simple string, but I've been trying to figure out how to get a negative lookbehind working on that apostrophe to ensure that it is not preceded by a backslash... Any ideas? - JMT

Original source