PHP regular expression match text not within strings

Question

I have content like foo == 'bar test baz' and test.asd = "buz foo". I need to match the "identifiers", the ones on the left that are not within double/single quotes. This is what I have now:

preg_replace_callback('#([a-zA-Z\\.]+)#', function($matches) {
    var_dump($matches);
}, $subject);

It now matches even those within strings. How would I write one that does not match the string ones?

Another example: foo == 5 AND bar != 'buz' OR fuz == 'foo bar fuz luz'. So in essence, match a-zA-Z that are not inside strings.

@k102: It's not that simple. I can't make up every possible variation, but the subject can vary a lot in structure. For example: `foo = 'bar' AND baz = 'foo'`. — Tower, Dec 08 '11 at 09:29

score 1 · Accepted Answer · edited May 23 '17 at 11:48

/^[^'"=]*/

would work on your examples. It matches any number of characters (starting at the start of the string) that are neither quotes nor equals signs.

/^[^'"=\s]*/

additionally avoids matching whitespace which may or may not be what you need.

Edit:

You're asking how to match letters (and possibly dots?) outside of quoted sections anywhere in the text. This is more complicated. A regex that can correctly identify whether it's currently outside of a quoted string (by making sure that the number of quotes, excluding escaped quotes and nested quotes, is even) looks like this as a PHP regex:

'/(?:
 (?=      # Assert even number of (relevant) single quotes, looking ahead:
  (?:
   (?:\\\\.|"(?:\\\\.|[^"\\\\])*"|[^\\\\\'"])*
   \'
   (?:\\\\.|"(?:\\\\.|[^"\'\\\\])*"|[^\\\\\'])*
   \'
  )*
  (?:\\\\.|"(?:\\\\.|[^"\\\\])*"|[^\\\\\'])*
  $
 )
 (?=      # Assert even number of (relevant) double quotes, looking ahead:
  (?:
   (?:\\\\.|\'(?:\\\\.|[^\'\\\\])*\'|[^\\\\\'"])*
   "
   (?:\\\\.|\'(?:\\\\.|[^\'"\\\\])*\'|[^\\\\"])*
   "
  )*
  (?:\\\\.|\'(?:\\\\.|[^\'\\\\])*\'|[^\\\\"])*
  $
 )
 ([A-Za-z.]+) # Match ASCII letters/dots
)+/x'

An explanation can be found here. But probably a regex isn't the right tool for this.

Do you have an idea that could work in the just-added example in the question? — Tower, Dec 08 '11 at 09:45
@rFactor Your edited question is unclear. Please provide also the desired output. — FailedDev, Dec 08 '11 at 09:49
This is a bit more difficult, especially because quotes can be escaped or contain other quotes like `"2\" by 4\""`, `"Don't"` and such. — Tim Pietzcker, Dec 08 '11 at 09:50

score 1 · Answer 2 · answered Dec 08 '11 at 09:08

1

You could also try this:

preg_match_all('/[\w.]+(?=(?:[^\'"]|[\'"][^\'"]*["\'])*$)/', $subject, $result, PREG_PATTERN_ORDER);
for ($i = 0; $i < count($result[0]); $i++) {
    # Matched text = $result[0][$i];
}

To match all letters, digits and _ and dots outside your quotes. You can extend your allowable characters by adding them into [\w.]

answered Dec 08 '11 at 09:08

FailedDev

26,680
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I'm doing a preg_replace_callback, because I need to replace those matches. If I just do a match, how do I know I'm replacing the right content (e.g., `foo = 'foo'` should match the first `foo` and replace it with my custom stuff, but it should not affect the one in the strings)? – Tower Dec 08 '11 at 09:46
@rFactor You could use the same regex with preg_replace_callback. Point is, with your edited question, it is unclear to me which characters you want to catch. – FailedDev Dec 08 '11 at 09:52

Marijn van Vliet · Answer 3 · 2011-12-08T14:25:46.157

1

The trick I use here is to force the regex to branch whenever it encounters a quote, then later on we ignore this branch.

$subject = <<<END
foo == 'bar test baz' and test.asd = "buz foo"
foo == 5 AND bar != 'buz' OR fuz == 'foo bar fuz luz'
END;

$regexp = '/(?:["\'][^"\']+["\']|([a-zA-Z\\.]+\b))/';

preg_replace_callback($regexp, function($matches) {;
    if( count($matches) >= 2 ) {
        print trim($matches[1]).' ';
    }
}, $subject);

// Output: 'foo and test.asd foo AND bar OR fuz '

The main part of the regexp is

(?: anything between quotes | any word consisting of a-zA-Z )

edited Dec 08 '11 at 14:25

answered Dec 08 '11 at 09:11

Marijn van Vliet

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Sorry if I was too imprecise. I believe that does not exactly match the case I added in the question? Or maybe I don't fully understand the regex, but by the look of it, it assumes there are no characters besides a-z and quotes? – Tower Dec 08 '11 at 09:41
I've modified my answer to only match the a-zA-Z case. Ask if the regexp is unclear to you. – Marijn van Vliet Dec 08 '11 at 14:21

PHP regular expression match text not within strings

3 Answers3