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.

Answer 1

/^[^'"=]*/

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.