Question

我正在使用像这样的Array_sum

    <?php
     $totalAmount = $db->prepare('SELECT
     a.proId, a.userId,
     b.id, b.pPrice
     FROM purchaseshistory AS a
     INNER JOIN products AS b ON(a.proId=b.id)
     WHERE a.userId=?');
     $totalAmount->bind_param('i', $cus['cId']);
     $totalAmount->execute();
          $totalAmount->bind_result($proId, $userId, $id, $pPrice);
$totalAmount->store_result();
   while ($totalAmount->fetch()) {
     $sum = 0;
     $amount = $pPrice;
     $amount = is_array($amount) ? $amount : array($amount);
          foreach ($amount as $item => $value) {
           $sum += $value;
}
print $sum;
}
?>

并试图像这样做

while ($totalAmount->fetch()) {
       $amount = array($pPrice);
       print array_sum($amount) ;
}

相同的结果我得到像200150200这样的数字

Answer 1

$amount = [];
while ($totalAmount->fetch()) {
       $amount[] = $pPrice;
}
echo array_sum($amount);

你没有向数组添加元素..上面的代码是，并且应该按预期工作

旧版PHP版本的更新

$amount = array();
while ($totalAmount->fetch()) {
       $amount[] = $pPrice;
}
echo array_sum($amount);

Answer 2

另一种选择（不使用array_sum（））： -

$sum = 0;
while ($totalAmount->fetch()) {
  $sum += $pPrice;
}
echo $sum;

array_sum返回没有sum的原始数字 - php

2 个答案: