php - 当我尝试插入到表中时，如何修复当前位置的错误？

Question

我想在用户售出商品后执行此操作，但如果我的最终库存少于我的最小库存，那么它会从数据库中在我的表上插入一条通知，但在使用 node.js 推送该通知之后，但是当我尝试插入售后表时，显示这样的错误，我应该如何修复它？

public function concretar_venta(){
            if($this->sale->checa_carrito_vacio($this->session->carrito)){
                $total = $this->input->post("total", TRUE);
                $cantidad_pagada = $this->input->post("cantidad_pagada", TRUE);
                $cambio = $cantidad_pagada - $total;
                if($this->sale->concretar_venta($this->session->carrito, $total, $cantidad_pagada, $cambio)){
                    $this->json(array('success' => 'The sale was successfully made'));
                }
                else{
                    $this->json(array('error' => 'There was an error making the sale, please try again'));
                }

                $this->session->carrito = $this->sale->checar_existe_carrito();
                $array = $this->sale->get_all_cart($this->session->carrito);
                $product_id = array();
                foreach ($array as $key => $value) {
                    $product_id[] = $value['id'];
                }

                $this->notification->addNotification('low stock', $product_id, $this->session->log['id'], 'low stock');

                /*if ($product->stock <= 8) {
                    $this->notification->addNotification('low stock', $product_id, $this->session->log['id'], 'low stock');
                } else {
                    # code...
                }*/

            }
            else{
                $this->json(array('error' => 'The cart is empty'));
            }
        }

模型通知:

public function addNotification($message, $product_id, $user_id, $type = ''){
        $types = array('new' => 0, 'pending' => 1, 'low stock' => 2);
        if (isset($types[$type]) === false) {
            throw new \InvalidArgumentException('Value for third parameter must be one of new, pending, or low stock.');
        }
        $type = $types[$type];
        $timestamp = time();
        $query = "SELECT COUNT(*) AS notificationCount FROM storelte_notifications WHERE product_id IN ? AND type = ? ";
        $previousNotification = $this->db->query($query, array($product_id, $type))->result_array();
        if ($previousNotification[0]['notificationCount'] == 0) {
            $sql = "INSERT INTO storelte_notifications (message,type,product_id,user_id,timestamp) VALUES(?, ?, ?, ?, ?)";
            try {
                foreach ($product_id as $pid) {
                    if (!$this->db->query($sql, array($message, $type, $pid, $user_id, $timestamp))) {
                        return false;
                    }
                }
                return true;
            } catch (Exception $e) {

            }
        }else{
            return true;
        }
    }

错误输出:

Error Number: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ') AND type = 2' at line 1 SELECT COUNT(*) AS notificationCount FROM storelte_notifications WHERE product_id IN () AND type = 2 Filename: C:/xampp/htdocs/storelte/system/database/DB_driver.php Line Number: 691

Answer 1

IN() 需要 checkin 一串 Product_ids，因此您需要将 Product_ids 数组转换为字符串

这两行之间:

 $timestamp = time();
 $query = "SELECT COUNT(*)...

添加

 $timestamp = time();
 $product_id = implode(',',$product_id);//add this line
 $query = "SELECT COUNT(*)...

http://php.net/manual/en/function.implode.php