迭代数组中的前30个键

Question

I have an array stored into a variable called $data

["data"]=>
  array(413) {
    [0]=>
    object(stdClass)#254 (7) {
      ["Energy"]=>
      string(7) "44555"
      ["Closing"]=>
      string(10) "102"
    }
    [1]=>
    object(stdClass)#260 (7) {
      ["Energy"]=>
      string(7) "2522"
      ["Closing"]=>
      string(10) "854"
    }

And I obtain an arithmetic mean for all the Closing values like this:

// initialize sum and total
$sum = 0;
$total = 0;

foreach ($data->data as $obj) {
    if (isset($obj->Closing)) {   // verify that the current object has a Closing
        $sum += $obj->Closing;    // add it to the sum
        $total++;                  // increment the count
    }
}
echo $sum / $total;                // display the average

The problem is that now I will only need the first 30 key from the array, but I don't know how to do that, can anyone help me please? I have tried with a for loop but it doesn't do the trick. Any ideas? Thank you all in advance for your time and help.

Answer 1

Using your second piece of code :

$sum = 0;
$total = 0;

foreach ($data->data as $obj) {
    if (isset($obj->Closing) && $total < 30) {   // Closing? + numOfValues < 30 ?
        $sum += $obj->Closing;
        $total++;
    }
}

// Display the average
echo $sum / $total;

Adding && $total < 30 should do the trick since you count how many items you add.

---

You can break out of the for if you don't want to do extra loops with a single extra condition :
(thanks Adam for pointing this out)

foreach ($data->data as $obj) {
    if (isset($obj->Closing) && $total < 30) {
        // ...
    }
    else if ($total >= 30) break;
}