phpregexreplacepreg-replacepreg-match

Replace literal string in text if another literal string doesn't exist in the text

I have these 3 strings:

$string = "his";
$str1 = "hi";
$str2 = "s";

So what I want to do is a regex that looks for hi and replaces it. But! if there is s in the string it won't be replaced. Like this.

preg_replace('/'.$str1.'^['.$str2.']/', "replace it with this", $string);

It's not working! (of course not, regex isn't my thing!)

As I said, I don't get this with regex. I want to find $str1 and if $str2 isn't in the string it won't be replaced.

Solution 
  • $str = 'his';

$s1 = 'hi';
$s2 = 's';

$result = preg_replace('~' . preg_quote($s1) . '(?!' . preg_quote($s2) . ')~', 'replace with this', $str);
                      // ~hi(?!s)~
                      // this regex means:
                      //   "hi" string followed by anything but "s"

var_dump($result);

    Live examples:

    1. http://ideone.com/XjX9n3
    2. http://ideone.com/U2JdkL