how to replace arrays by pointers from other array

Question

I got an array with values splits to 5 arrays with groups of 3 values, then I got more group of array of pointers, what I'm trying to do is replacing the pointed array with other value.

what I got is :

$a = array(["4", "3", "5"], ["4", "8", "11"], ["4", "11", "5"], ["4", "5", "5"], ["4", "4", "9"]);

$b = array(12); // the value to be placed

$pointers = array([1,2],[2,2],[2,1]);

foreach($pointers as $point)
{
    $a1 = array_replace($a[$point[0]], array($point[1] => $b ));
}

Expected results:

["4", "3", "5"], ["4", "8", "12"], ["4", "12", "12"], ["4", "5", "5"], ["4", "4", "9"]

What I need is to keep the array structure as is and just replace the pointed values to be 12.

I'm not sure its the right way.

You're overwriting `$a1` each time through the loop. The result will just be the replacement from the last element of `$pointer`. — Barmar, Jul 31 '22 at 20:06
Why is `$b` an array? That will create a nested array in the replacement. — Barmar, Jul 31 '22 at 20:08
Related: https://stackoverflow.com/a/27930028/2943403 and https://stackoverflow.com/q/31357558/2943403 — mickmackusa, Aug 02 '22 at 10:55

score 1 · Accepted Answer · answered Jul 31 '22 at 20:11

You don't need to use array_replace(), just assign using the indexes in $pointers.

$a = array(["4", "3", "5"], ["4", "8", "11"], ["4", "11", "5"], ["4", "5", "5"], ["4", "4", "9"]);
$b = 12; // the value to be placed
$pointers = array([1,2],[2,2],[2,1]);

foreach($pointers as [$p1, $p2])
{
    $a[$p1][$p2] = $b;
}
print_r($a);

how to replace arrays by pointers from other array

1 Answers1