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Чайник
Asked:2020-11-03 22:26:08 +0000 UTC2020-11-03 22:26:08 +0000 UTC

在 PDO 上更新数据库中的数据

  • 772

朋友告诉我怎么做！我使用 Ajax 将数据发送到服务器以添加到数据库中，在服务器上使用 UPDATE 的常用代码。

我想不出一个通用代码，这样我就不会为同一张表的不同更新请求编写新代码。

例如，如果我只更新成本，那么这是一个UPDATE；如果更新成本和描述，那么这已经不同了，但是可以有很多选择。

//обновление заказа в базе денных
    public function update_order($order, $pdo){

        $sql = $pdo->prepare("UPDATE `orders` SET 
            `company` = :company, `order_name` = :order_name, `status` = :status, `dead_line` = :dead_line, `cost` = :cost, `payment` = :payment, `description` = :description, `bill` = :bill, `act` = :act, `contract` = :contract WHERE `id` = :id");
        $sql->execute([

            ':id' => $order['id'],
            ':company' => $order['company'],
            ':order_name' => $order['order_name'],
            ':status' => $order['status'],
            ':dead_line' => $order['dead_line'],
            ':cost' => $order['cost'],
            ':payment' => $order['payment'],
            ':description' => $order['description'],
            ':bill' => $order['bill'],
            ':act' => $order['act'],
            ':contract' => $order['contract']

        ]);
        return $order['dead_line'];
    }//end addOrder
php

1 个回答

  1. Best Answer
    //обновление заказа в базе денных
public function update_order($order, $pdo){
    $query_list = [];
    $execute_list = ["id" => $order["id"]];
    $fields = ["company", "order_name", ];
    foreach ($fields as $field) {
        if (!empty($order[$field]) {
            $query_list[] = "`$field` = :$field";
            $execute_list[":$field"] = $order[$field];
        }
    }

    $sql = $pdo->prepare("UPDATE `orders` SET " . implode(", ", $query_list) . " WHERE `id` = :id");
    $sql->execute(execute_list);
    return $order['dead_line'];
}//end addOrder
    • 0

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