创建第三个数组，即array1 + array2

<?php

/*ARRAY 1*/

$numbers = [];

while (count($numbers) < 100) {
    $numbers[] = random_int(1, 6);

}

implode($numbers);

echo "<b>Array nummer 1:</b> ";

foreach ($numbers as $number) {
    echo $number . " ";

}

echo "<br><br><b>Array nummer 2:</b> ";

/*ARRAY 2*/

$numbers2 = [];

while (count($numbers2) < 100) {
    $numbers2[] = random_int(1, 6);

}

implode($numbers2);

foreach ($numbers2 as $number2) {
    echo $number2 . " ";
}

/*ARRAY 3 is not right*/

echo "<br><br>";

print_r ($numbers+$numbers2);



?>

I am practising php now and i was wondering how to count up array's like this: array1 = 1,2,4; array2 = 3,6,8;

so array 3 must be = 4,8,12

I was searching online but couldn't find answers on this so that's why i am asking here.

I suspect you are trying to achieve this:

$numbers3 = array();
for ($i=0; $i < count($numbers) && $i < count($numbers2); $i++) { 
    $numbers3[] = $numbers[$i] + $numbers2[$i];
}

The + operator doesn't work like making the sum of the elements on both arrays.

The + operator (on arrays) returns the right-hand array appended to the left-hand array; for keys that exist in both arrays, the elements from the left-hand array will be used, and the matching elements from the right-hand array will be ignored. See: http://php.net/manual/en/language.operators.array.php

Also, the implode doesn't work by reference. The result of the implode should be assigned to a variables. The implode($numbers2); doesn't makes anything (it executes, but you don't save the result)