replace multiple spaces in a string with a single space

Your regex replaces any number of spaces (including zero) with a space. You should only replace two or more (after all, replacing a single space with itself is pointless):

$s = preg_replace("/ {2,}/", " ", $x);

// Your input
$str = "Hjhajhashsh dwddd dddd sss   ddd wdd ddcdsefe xsddd   scdc yyy5ty    ewewdwdewde           wwwe ddr3r dce eggrg               vgrg fbjb   nnn  bh jfvddffv mnmb   weer ffer3ef f4r4 34t4 rt4t4t 4t4t4t4t    ffrr  rrr  ww w w ee3e iioi   hj   hmm  mmmmm mmjm lk ;’’ kjmm  ,,,, jjj hhh  lmmmlm m mmmm lklmm jlmm m";
        echo $str.'<br>'; 

        $output = preg_replace('!\s+!', ' ', $str); // Replace multispace with sigle.

        echo $output;

What I usually do to clean up multiple spaces is:

while (strpos($x, '  ') !== false) {
   $x = str_replace('  ', ' ', $x);
}

Conditions/hypotheses:

1. strings with multiple spaces are rare
2. two spaces are by far more common than three or more
3. preg_replace is expensive in terms of CPU
4. copying characters to a new string should be avoided when possible

Of course, if condition #1 is not met, this approach does not make sense, but it usually is.

If #1 is met, but any of the others is not (this may depend on the data, the software (PHP version) or even the hardware), then the following may be faster:

if (strpos($x, '  ') !== false) {
   $x = preg_replace('/  +/', ' ', $x); // i.e.: '/␣␣+/'
}

Anyway, if multiple spaces appear only in, say, 2% of your strings, the important thing is the preventive check with strpos, and you probably don't care much about optimizing the remaining 2% of cases.