Question

if(preg_match("/(.*(<code>).*(<\/code>).*)*/", $string))

I was trying many hours already, but I can’t make it work. I want that if user formatted text correctly like:

(any_string*<code>any_string*</code>any_string*)*

Only then I would format text with * meaning empty string or many times. What’s wrong with my expression?

edit: I want to match lalala text <code>dlalala lala code</code> lalalal. If it’s lala <code> lalala or lalal </code> <code> lalala <code>alala then we don’t want to match it.

Answer 1

This works with the 2 given test cases:

$arr = array('lalala text <code>dlalala lala code</code> lalalal.',
             'lala <code> lalala or lalal </code> <code> lalala <code>alala');
foreach ($arr as $str) {
    echo "$str\n";
    if (preg_match('#^(?<!<code>).*<code>.*?</code>(?!.*<code>)#', $str)) {
        echo "===> Match\n";
    } else {
        echo "===> Not match\n";
    }
}

output:

lalala text <code>dlalala lala code</code> lalalal.
===> Match
lala <code> lalala or lalal </code> <code> lalala <code>alala
===> Not match

Some explanation about the regex:

#           : regex delimiter
^           : begining of string
  (?<!      : start negative lookbehind
    <code>  : literally <code>
  )         : end of lookbehind
  .*        : any char any number of time
  <code>    : literally <code>
  .*?       : any char any number of time not greedy
  </code>   : literally </code>
  (?!       : start negative lookahead
    .*      : any char any number of time
    <code>  : literally <code>
  )         : end of lookahead
#           : regex delimiter

You can find some usefull informations about lookaround here