[FIXED] How can I replace all matches with preg_replace only when not inside quotes?

Issue

can someone help me to solve this problem?

I have this string:

$app.url = $cfg.url ? $cfg.url : 'domain.ext'
    
    

What I need is convert it to this:

$app[url] = $cfg[url] ? $cfg[url] : 'domain.ext'

What I try:

$input = $app.url = $cfg.url ? $cfg.url : 'domain.ext';
        preg_replace('/(\.)(\w+)/', '[$2]', $input);

The result of above:

$app[url] = $cfg[url] ? $cfg[url] : 'domain[ext]'

I also try with:

$input = $app.url = $cfg.url ? $cfg.url : 'domain.ext';
            preg_replace('/(!\')(\.)(\w+)/', '[$2]', $input);

Result above:

$app.url = $cfg.url ? $cfg.url : 'domain.ext'

But no satisfactory result.

Thanks in advance.

Solution

You can use

'[^\\']*(?:\\.[^\\']*)*'(*SKIP)(*F)|\.(\w+)

See the regex demo.

In PHP, you can use

preg_replace("/'[^\\\\']*(?:\\\\.[^\\\\']*)*'(*SKIP)(*F)|\.(\w+)/s", '[$1]', $input)

Pattern details

• '[^\\']*(?:\\.[^\\']*)*'(*SKIP)(*F) - a single quoted string literal pattern:
  • ' - a ' char
  • [^\\']* - zero or more chars other than ' and \
  • (?:\\.[^\\']*)* - zero or more sequences of any escaped char and then zero or more chars other than \ and '
  • ' - a ' char
  • (*SKIP)(*F) - fails the match and triggers next match search from the failure position
• | - or
• \.(\w+) - a dot and then one or more word chars captured into Group 1.

Answered By - Wiktor Stribiżew

Answer Checked By - Pedro (PHPFixing Volunteer)