Xcode 8 Swift 3.0：解析错误 - 无法读取 PHPMailer 代码

我在 xcode 8 上有问题。

在 PHP 中，我使用 PHPMailer 发送电子邮件。我的 PHP 代码如下。

发送.php

<?php
    require 'database/connect.php';
    global $connect;
    date_default_timezone_set('Etc/UTC');
    require 'PHPMailer-master2/PHPMailerAutoload.php';

    if ( isset($_POST['data1']) && isset($_POST['data2']))
    {
        $data1 = $_POST['data1'];
        $data2 = $_POST['data2'];

        $sql        = "SELECT * FROM table WHERE data1 = '$data1' AND data2='$data2'";
        $result     = mysqli_query($connect, $sql);
        if ($result && mysqli_num_rows($result) > 0)
        {
            while ($row = mysqli_fetch_array($result)){
            }

            $output = array('message' => '1');
            echo json_encode($output);
            $add = "INSERT INTO table (data1, data2)
                               VALUES ('$data1','$data2')
            ";
            $run = mysqli_query($connect,$add);
            $mail = new PHPMailer;                             
            $mail->isSMTP();                                       
            $mail->Host = 'smtp.gmail.com';  
            $mail->SMTPAuth = true;                                
            $mail->Username = 'gmail.com';               
            $mail->Password = '******';                            
            $mail->SMTPSecure = 'tls';                            
            $mail->Port = 587;                                     
            $mail->setFrom('[email protected]', 'sender');  
            $mail->addAddress('[email protected]','receiver');
            $mail->isHTML(true);                                  
            $mail->Subject = 'Test';
            $mail->Body    = 'Test';
            $mail->AltBody = 'Test';
        if(!$mail->send()) {
            echo json_encode([
                'status' => false,
                'message' => 'Message could not be sent. Error: ' . $mail->ErrorInfo
            ]);
        } else {
            $status   = array();
            $status[] = array('status' => '1');
        }

        $output = array('message' => '1', 'status' => $status);
        echo json_encode($output);
        exit();
            // End sending email
            exit();

        mysqli_free_result($result);
        }
        else {}
    }
?>

我设法使用上面的代码将数据发送到服务器并发送电子邮件给接收者。

我现在面临的唯一问题是在 xcode 中。它说：

解析错误：无法读取数据，因为它的格式不正确。

Xcode 无法读取 PHP 文件中的 PHPMailer 代码，这导致我的 swift 3.0 代码执行Catch语句而不是message == '1'语句。我的快速代码如下。

后.swift

@IBAction func sendApplyMovement(_ sender: Any) {
            let url             = URL(string: "http://localhost/send.php")
            let session         = URLSession.shared
            let request         = NSMutableURLRequest(url: url! as URL)
            request.httpMethod  = "POST"
            let valueToSend     = "data1=&data2"
            request.httpBody    = valueToSend.data(using: String.Encoding.utf8)

            let myAlert         = UIAlertController(title: "Confirm", message: "Sure ?", preferredStyle: UIAlertControllerStyle.alert)
            let cancel          = UIAlertAction(title: "Cancel", style: UIAlertActionStyle.default, handler: nil)
            let okaction        = UIAlertAction(title: "Yes", style: UIAlertActionStyle.default, handler:
                {
                    action in

            let task = session.dataTask(with: request as URLRequest, completionHandler: {
                (data, response, error) in
                if error != nil {
                    return
                }
                else {
                            do {
                                if let json = try JSONSerialization.jsonObject(with: data!) as? [String: String]
                                {
                                    DispatchQueue.main.async {
                                        let message   = Int(json["message"]!)
                                        let status    = Int(json["status"]!)

                                        if(message == 1){
                                            if(status == 1){
                                                print("Success")
                                                let myViewController:ViewController = self.storyboard!.instantiateViewController(withIdentifier: "ViewController") as! ViewController
                                                let appDelegate = UIApplication.shared.delegate as! AppDelegate
                                                let navigationController = UINavigationController.init(rootViewController: myViewController)
                                                appDelegate.window?.rootViewController = navigationController
                                                appDelegate.window?.makeKeyAndVisible()

                                                let myAlert = UIAlertController(title: "Success!", message: "Sent !", preferredStyle: UIAlertControllerStyle.alert)
                                                myAlert.addAction(UIAlertAction(title: "Okay", style: UIAlertActionStyle.default, handler: nil))
                                                navigationController.present(myAlert, animated: true, completion: nil)
                                                return
                                            }

                                        }
                                        else {return}
                                    }
                                }
                            }
                            catch let parseError { print("Parse error: \(parseError.localizedDescription)") }
                }
            })
            task.resume()

            }
            )
            myAlert.addAction(okaction)
            myAlert.addAction(cancel)
            self.present(myAlert, animated: true, completion: nil)
        }
    }

有什么我需要修改才能使它工作的吗？