replace last seven characters of string

Question

I have a string and I want to replace the last 7 charators of the string with "#". For example I have "MerryChristmasu87yujh7" I want to replace "87yujh7" with seven "#######". So, the final string will be "MerryChristmasu#######".

I have tried the follow code but it returns "MerryChristmasu#######1". It does not convert all seven ending characters.

$string = "MerryChristmasu87yujh7";
$match = substr($string, -7, -1);
$result = str_replace($match, "#######", $string);

Answer 1

Should be...

$match = substr($string, -7);

... without the final -1. But in fact, it's far better done with...

$result = substr($string, 0, -7) . str_repeat('#', 7);

... or, more generic:

$coverWith = function($string, $char, $number) {
  return substr($string, 0, -$number) . str_repeat($char, $number);
};

Answer 2

$cuttedString = substr("your string", -7);

this should do the job.