How can regex ignore escaped-quotes when matching strings?

问题

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

回答1:

<?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\\'

回答2:

Here's my solution with test cases:

/.*?'((?:\\\\|\\'|[^'])*+)'/

And my (Perl, but I don't use any Perl-specific features I don't think) proof:

use strict;
use warnings;

my %tests = ();
$tests{'Case 1'} = <<'EOF';
$var = 'My string';
EOF

$tests{'Case 2'} = <<'EOF';
$var = 'My string has it\'s challenges';
EOF

$tests{'Case 3'} = <<'EOF';
$var = 'My string ends with a backslash\\';
EOF

foreach my $key (sort (keys %tests)) {
    print "$key...\n";
    if ($tests{$key} =~ m/.*?'((?:\\\\|\\'|[^'])*+)'/) {
        print " ... '$1'\n";
    } else {
        print " ... NO MATCH\n";
    }
}

Running this shows:

$ perl a.pl
Case 1...
 ... 'My string'
Case 2...
 ... 'My string has it\'s challenges'
Case 3...
 ... 'My string ends with a backslash\\'

Note that the initial wildcard at the start needs to be non-greedy. Then I use non-backtracking matches to gobble up \\ and \' and then anything else that is not a standalone quote character.

I think this one probably mimics the compiler's built-in approach, which should make it pretty bullet-proof.

回答3:

/.*'([^'\\]|\\.)*'.*/

The parenthesized portion looks for non-apostrophes/backslashes and backslash-escaped characters. If only certain characters can be escaped change the \\. to \\['\\a-z], or whatever.

回答4:

Via negative look behind:

/
.*?'              #Match until '
(
 .*?              #Lazy match & capture of everything after the first apostrophe
)    
(?<!(?<!\\)\\)'   #Match first apostrophe that isn't preceded by \, but accept \\
.*                #Match remaining text
/

回答5:

Regex reg = new Regex("(?<!\\\\)'(?<string>.*?)(?<!\\\\)'");

回答6:

This is for JavaScript:

/('|")(?:\\\\|\\\1|[\s\S])*?\1/

it...

matches single or double quoted strings
matches empty strings (length 0)
matches strings with embedded whitespace (\n, \t, etc.)
skips inner escaped quotes (single or double)
skips single quotes within double quotes and vice versa

Only the first quote is captured. You can capture the unquoted string in $2 with:

/('|")((?:\\\\|\\\1|[\s\S])*?)\1/

来源：https://stackoverflow.com/questions/1060616/how-can-regex-ignore-escaped-quotes-when-matching-strings

标签

regex

escaping

negative-lookbehind