random（）函数返回1到10之间的整数

how to Ensure that it can not return one of the values ​​in the array $ excluded.

This is my code

$exclus=array(1,4);
function ramdom(){
   $resultat=rand(1,10);
   return $resultat;
}

echo "Random Num : ".random()."
";

try this:

function randWithout($from, $to, array $exceptions) {
    sort($exceptions); // lets us use break; in the foreach reliably
    $number = rand($from, $to - count($exceptions)); // or mt_rand()
    foreach ($exceptions as $exception) {
        if ($number >= $exception) {
            $number++; // make up for the gap
        } else /*if ($number < $exception)*/ {
            break;
        }
    }
    return $number;
}

and call it:

$exclus=array(1,4);
$random_number = randWithout(0,10,$exclus);

Here is an online demo of the code below: https://eval.in/87096

function ramdom() {
   $resultat = rand(1,10);    
   $exclus = array(1,4);
   if(in_array($resultat,$exclus)) {
      return ramdom();
   }
   else {
     return $resultat;
  }
}

Use array_search :

$exclus=array(1,4);
function random(){
   $resultat=rand(1,10);
 return array_search($resultat,$exclus) ? ramdom() : $resultat
}

<?php
$exclus = array(1,4);
function ramdom()   {
    $resultat = rand(1,10);
    while(in_array($resultat, $exclus)) {
        $resultat = rand(1,10);
    }
    return $resultat;
}

?>

Should work, but I didn't test it...