提交表单后,我收到一个索引数组作为 SQL 查询结果,目标是将特定于列的值与之前收集的结果相加。
示例:
// 1st query result
$product_amount = [0.36, 0.14, 0.42]
// 2nd query result
$product_amount = [0.28, 0.12, 0.40]
// n-th query result
$product_amount = [0.16, 0.14, 0.42]
// the desired sum
$total = [0.80, 0.40, 1.24]
如何做?
创建数组的代码:
function doseCount($num){
return ($num/100) * intval($_POST['portion']); //getting multiplier for productAmount() function
}
function productAmount(){
global $link; //connection to the MySQL server
if(isset($_POST['product'])){
$product_amount_query = 'SELECT va.vitamin_a, va.vitamin_b1, va.vitamin_b2, va.vitamin_b3, va.vitamin_b5,
va.vitamin_b6, va.vitamin_bc_b9, va.vitamin_b12, va.vitamin_c, va.vitamin_d, va.vitamin_e, va.biotin, va.vitamin_k,
va.flavonoids, va.lipoic_acid FROM products_list p, vitamins_amount va WHERE p.id = va.product_id AND p.name = ?';
$product_amount_stmt = mysqli_prepare($link, $product_amount_query);
mysqli_stmt_bind_param($product_amount_stmt, "s", $_POST['product']);
mysqli_stmt_execute($product_amount_stmt);
$result = mysqli_stmt_get_result($product_amount_stmt);
$product = mysqli_fetch_array($result, MYSQLI_NUM);
return $product_final_result = array_map('doseCount', $product);
mysqli_stmt_close($product_amount_stmt);
mysqli_close($link);
}
}
$product_amount = productAmount();
提交后,
$product_amount
数组获得新数据,并且此新数据必须添加到之前积累的数据中。
您可以通过组合 array_map 和 array_sum 来实现此目的。如果您使用 PHP 5.6 或更高版本(您应该使用 PHP 5.6 或更高版本),您可以编写如下内容:
$total = array_map(function (...$vals) {
return array_sum($vals);
}, ...$results);
假设
$results
包含所有 $product_amount
数组。
如果您想增量构建
$total
而不是一次传递所有查询结果,您可以这样做:
// Initialize $total array
$total = [0, 0, 0];
// Update $total for each query
$total = array_map(function (...$vals) {
return array_sum($vals);
}, $total, $product_amount);
这个问题可以通过正确使用会话来解决:
if (empty($_SESSION['product_amount'])){
$_SESSION['product_amount'] = $product_amount;
}else{
foreach ($_SESSION['product_amount'] as $key => $value) {
$_SESSION['product_amount'][$key] += $product_amount[$key];
}
}
特别感谢 Mawia HL 的帮助。
我有一些疑虑和建议:
global
来获取所需值。明确地将所需数据传递给您的函数。$_POST
——这些值应显式传入,以便您的代码可以更轻松地进行单元测试。$portion
的计算很小且无条件,我可能会直接将该步骤构建到 productAmount()
中。array_sum()
的映射调用,如果您愿意,甚至可以将浮点值截断为一定的小数位数。代码:(PHPize演示)
function productAmount($mysqli, $productName, $portion): array
{
$sql = '
SELECT va.vitamin_a,
va.vitamin_b1,
va.vitamin_b2,
va.vitamin_b3,
va.vitamin_b5,
va.vitamin_b6,
va.vitamin_bc_b9,
va.vitamin_b12,
va.vitamin_c,
va.vitamin_d,
va.vitamin_e,
va.biotin,
va.vitamin_k,
va.flavonoids,
va.lipoic_acid
FROM products_list p
JOIN vitamins_amount va ON p.id = va.product_id
WHERE p.name = ?
';
$stmt = $mysqli->prepare($sql);
$stmt->execute([$productName]);
return array_map(
fn(int $amount) => ($amount / 100) * $portion,
$stmt->get_result()->fetch_array(MYSQLI_NUM)
);
}
$POSTs = [
['product' => 'goodinya', 'portion' => 3],
['product' => 'healthy1', 'portion' => 2],
['product' => 'vitastic', 'portion' => 4],
];
$result = [];
foreach ($POSTs as ['product' => $product, 'portion' => $portion]) {
$result[] = productAmount($mysqli, $product, $portion);
}
var_export(
array_map(
fn(...$rows) => sprintf('%0.2f', array_sum($rows)),
...$result
)
);