更新嵌套数组的多个值

how to I update if let say user input chemistry and biology that they want discount price. How do I go to the nested array of user, itemprice to update its value?

    [name] => xxxx
    [phone] => xxxxx
    [email]xxxxx
    [itemprices] => Array ( [0] => 1.00 [1] => 1.00 [2] => 1.00)
    [iteminfo] => Array ( [0] => Chemistry [1] => Biology [2] => Mathematics) 
    )

I've tried with the solution below, but when I update only chemistry, it will update the biology and mathematics's itemprice together.

Why is that so?

$subject = 'Chemistry';
$index = array_search($subject, $user->iteminfo);
if (false !== $index) {
  $user->itemprices[$index] = $newvalue;
}

It works like a charm I rewrite it, you can try it out

$user = (object) array(
    'name' => 'xxxx',
    'phone' => 'xxxxx',
    'itemprices' => Array (1.00, 1.00, 1.00),
    'iteminfo' => Array ('Chemistry', 'Biology', 'Mathematics') 
    );

echo "<pre>";
var_dump($user);
echo "</pre>";


$newvalue = 2.0;
$subject = 'Chemistry';

$index = array_search($subject, $user->iteminfo);
if (false !== $index) {

  $user->itemprices[$index] = $newvalue;

}

echo "<br><br><pre>";
var_dump($user);
echo "</pre>";

output

object(stdClass)#21 (4) {
  ["name"]=>
  string(4) "xxxx"
  ["phone"]=>
  string(5) "xxxxx"
  ["itemprices"]=>
  array(3) {
    [0]=>
    float(1)
    [1]=>
    float(1)
    [2]=>
    float(1)
  }
  ["iteminfo"]=>
  array(3) {
    [0]=>
    string(9) "Chemistry"
    [1]=>
    string(7) "Biology"
    [2]=>
    string(11) "Mathematics"
  }
}


object(stdClass)#21 (4) {
  ["name"]=>
  string(4) "xxxx"
  ["phone"]=>
  string(5) "xxxxx"
  ["itemprices"]=>
  array(3) {
    [0]=>
    float(2)
    [1]=>
    float(1)
    [2]=>
    float(1)
  }
  ["iteminfo"]=>
  array(3) {
    [0]=>
    string(9) "Chemistry"
    [1]=>
    string(7) "Biology"
    [2]=>
    string(11) "Mathematics"
  }
}

You are mixing Objects and Arrays, change $user->iteminfo to $user['iteminfo'] and $user->itemprices[$index] to $user['itemprices'][$index] and it would work properly.