如何求和特定数组项的值？

Question

I have a array like this:

$results = $stmt->fetchAll(PDO::FETCH_ASSOC);

/*
Array
(
    [0] => Array
        (
            [numb] => 10
            [range] => today
        )

    [1] => Array
        (
            [numb] => 5
            [range] => today
        )

    [2] => Array
        (
            [numb] => 5
            [range] => yesterday
        )

    [3] => Array
        (
            [numb] => 15
            [range] => in last week
        )

    [4] => Array
        (
            [numb] => 10
            [range] => in last week
        )

    [5] => Array
        (
            [numb] => 5
            [range] => in last week
        )

    [6] => Array
        (
            [numb] => 15
            [range] => in last month or more
        )
)
*/

Always there is 4 or less cases:

• today
• yesterday
• in last week
• in last month or more

And I'm trying to sum the numb item where range is today and there is another case in the $results array (or yesterday, or in last week, or in last month or more, or some of them, or all of them, doesn't matter .. Just there should be one more case except "today") .

How can I do that?

---

$score = null;
foreach($results as item) {
    if ( $item[range] == 'today' && /* another case exists */ ) {
        $score = item['numb'];
    } else {
        break;
    }
}

Note: I wrote that break because the items are sorted in the array. I mean always today 's items are either not exist or in the top of the array. So if that condition is FALSE then it also will be FALSE for the rest of items.

Answer 1

This ought to do it:

$score = 0;
foreach($results as $item) {
    if ( $item['range'] == 'today') {
        $score += item['numb'];
    }
}
var_dump($score); // The sum total of all the $result array's "numb" values

Answer 2

The other answers are missing a couple of keys he asked for in the question. He is looking to sum entries that are "today" if (and only if) there exist some entry in the array that isn't "today". While I agree the structure of the current array isn't ideal for this task, for the sake of simplicity, we will keep it and break it into a couple of steps

//first lets scan the array to see if there is something that is not "today"
$allTodays = true;
foreach ($result as $row) {
   if($row['range'] !== 'today') {
    $allTodays = false;
    break;
  }
}
// now if there is something that is not today
// sum the todays
$score = null;
if (!$allTodays) {
  foreach ($results as $item) {
    if ($item['range'] == 'today') {
        $score += item['numb'];
    } else {
        break;
    }
  }
}

You were pretty close I feel like.

Note: This may not be the most elegant solution, but it's a simple and straight-forward solution that should be easy to follow.

Answer 3

you can change the array structure in order to sum :

<?php
$sums = array();
// init
foreach($results as $item) {
     if($item['range'] == 'today') { // if you only want today add this condition
         if (!array_key_exists($item['range'], $sums)) {
             $sums[$item['range']] = 0;
         }
         $sums[$item['range']] += $item['numb'];
     }
}

// $sums['in last week'] will contain 30

Answer 4

If I understand you correctly, your task is a very common one - you want to group items first (based on 'range' field), then calculate sums.

The easiest solution, perhaps, is to store results for each range in an associative array:

$scores = [];

foreach ($results as $item) {
    if (! isset($scores[$item['range']])) {
        $scores[$item['range']] = $item['num'];
    } else {
        $scores[$item['range']] += $item['num'];
    }
}

print_r($scores);