PHP preg_replace remove all dot character from between square brackets

Question

How do I remove all . characters from between two square brackets in a string using preg_replace?

I'm trying to replace only between square brackets and not other dots in the string. This should have worked, but somehow just gives a blank string. How do I write the regex for this?

$str = '[city.name][city.state][city.mayor][city.mayor.name](city.name)';
$str = preg_replace('/\[.*?\]/','',$str);
echo $str;
// output
[cityname][citystate][citymayor][citymayorname](city.name)

Answer 1

You may use

'~(?:\G(?!^)|\[)[^][.]*\K\.~' # For [strings]
'~(?:\G(?!^)|<)[^<>.]*\K\.~'  # For <strings>

Or, to make sure there is a close ] there, add a (?=[^][]*]) lookahead:

'~(?:\G(?!^)|\[)[^][.]*\K\.(?=[^][]*])~' # For [strings]
'~(?:\G(?!^)|<)[^<>.]*\K\.(?=[^<>]*])~'  # For <strings>

See the regex demo and a regex demo with lookahead .

Details

• (?:\\G(?!^)|\\[) - a [ or end of the previous successful match
• [^][.]* - any 0+ chars other than [ , ] and .
• \\K - match reset operator
• \\. - a dot
• (?=[^][]*]) - a positive lookahead that requires a ] after any 0+ chars other than ] and [ immediately to the right of the current location.

PHP demo :

$str = '[city.name][city.state][city.mayor][city.mayor.name](city.name)';
echo preg_replace('~(?:\G(?!^)|\[)[^][.]*\K\.~', '', $str);

Answer 2

You could use \\G as in

(?:\G(?!\A)|\[)
[^\].]*\K\.

See a demo on regex101.com (mind the verbose mode).

---

Broken down, this says:

 (?: \\G(?!\\A) # match after the previous match (not the start) | # or \\[ # [ ) [^\\].]* # neither dot nor ] \\K # make the engine forget what's been matched before \\. # match a dot 

Answer 3

Use callback

$str = preg_replace_callback('/\[[^]]*\]/', function($m){
    return str_replace(".", "", $m[0]);
}, $str);