联合阵列的结果的总和

Question

Array 
(
    [0] => stdClass Object 
    (
     [id] => 1 
     [price] => 1.00 
     [session_price] => 1.00 
     [no_of_participants] => 1 
     [session_id] => 1 
     [coupon_id] => 0 
     [group_discount_id] => 0 
     [order_id] => 1 
     [created] => 2015-04-02 16:56:24 
     [modified] => 2015-04-02 16:56:24 
    ) 
    [1] => stdClass Object 
    (
     [id] => 2 
     [price] => 2.00 
     [session_price] => 2.00 
     [no_of_participants] => 2 
     [session_id] => 1 
     [coupon_id] => 1 
     [group_discount_id] => 1 
     [order_id] => 1 
     [created] => 2015-03-03 00:00:00 
     [modified] => 2015-03-03 00:00:00 
    ) 
    ) 

从上面我想总计no_of_participants（即1 + 2 = 3）。

Answer 1

通过与foreach阵列时，您环路，并把no_of_participants在一个变量$sum

$sum = 0; 
foreach($array as $object){ 
    $sum = $sum + $object->no_of_participants; 
} 

echo $sum; //Returns 3 

Answer 2

试试这个：

$sum = 0; 
foreach($datas as $data){ 
    $sum += $data->no_of_participants; 
} 

echo $sum; 

Answer 3

假设你在你的问题中定义的$对象：

然后按照下列步骤操作：

$array = json_decode(json_encode($object), true); 
$sum = 0; 
foreach($array as $val){ 
    $sum = $sum + $val->no_of_participants; 
} 
echo $sum; 

Answer 4

$sum = array_reduce(
    $myArrayOfObjects, 
    function($runningTotal, $record) { 
     $runningTotal += $record->no_of_participants; 
     return $runningTotal; 
    } 
);