php - 从 php 脚本返回数组到 Swift 时，JSON_encode 导致错误

Question

在 Xcode 中，我目前遇到以下错误:

Error Domain=NSCocoaErrorDomain Code=3840 "Garbage at end." UserInfo={NSDebugDescription=Garbage at end.}

当我向我的 php 脚本添加一个函数以从数据库中检索“社区”详细信息时，这才刚刚开始发生。这些详细信息存储在一个数组中。

下一行 echo json_encode($communities); 似乎是导致问题的原因，因为它在没有它的情况下运行良好。下面是完整的 userLogin.php 脚本。该行位于底部。

<?php

require ("Conn.php");
require ("MySQLDao.php");

$email = htmlentities($_POST["email"]);
$password = htmlentities($_POST["password"]);

$returnValue = array();


if(empty($email) || empty($password))
{
$returnValue["status"] = "error";
$returnValue["message"] = "Missing required field";
echo json_encode($returnValue);
return;
}

$secure_password = md5($password);

$dao = new MySQLDao();
$dao->openConnection();
$userDetails = $dao->getUserDetailsWithPassword($email,$secure_password);

if(!empty($userDetails))
{
$returnValue["status"] = "Success";
$returnValue["message"] = "User is registered";
echo json_encode($userDetails);
}else{

$returnValue["status"] = "error";
$returnValue["message"] = "User is not found";
echo json_encode($returnValue);
}

//once logged in run function to get list of communities

$communities = array();
$communities = $dao->getCommunities($email);
echo json_encode($communities);

$dao -> closeConnection();

?>

我已经在浏览器中测试了SQL函数，它返回了正确的值，输出如下:

[{"name":"EnclliffeT"},{"name":"OneWinner"},{"name":"Yesss"},{"name":"Treert"},{"name":"Westbrook"}]

我很确定问题出在 Swift 没有正确接收数组。

这是在用户登录时运行的 Swift 代码，它给出了错误:

 @IBAction func loginButtonTapped(_ sender: AnyObject)
    {

    let userEmail = userEmailTextField.text;
    let userPassword = userPasswordTextField.text;

    if (userPassword!.isEmpty || userEmail!.isEmpty) { return; }

// send user data to server side

    let myUrl = URL(string: "http://www.quasisquest.uk/KeepScore/userLogin.php");

    var request = URLRequest(url:myUrl!);
    request.httpMethod = "POST";
    let postString = "email=\(userEmail!)&password=\(userPassword!)";
    request.httpBody = postString.data(using: String.Encoding.utf8);

    let task = URLSession.shared.dataTask(with: request) { (data: Data?, response: URLResponse?, error: Error?) in
        DispatchQueue.main.async
            {

               if(error != nil)
                {

                    //Display an alert message
                    let myAlert = UIAlertController(title: "Alert", message: error!.localizedDescription, preferredStyle: UIAlertControllerStyle.alert);
                    let okAction = UIAlertAction(title: "OK", style: UIAlertActionStyle.default, handler:nil)
                    myAlert.addAction(okAction);
                    self.present(myAlert, animated: true, completion: nil)
                    return
                }

                do {

                    let json = try JSONSerialization.jsonObject(with: data!, options: .allowFragments) as? [String:AnyObject]


                    // retrieve login details and check to see if all ok

                    if let parseJSON = json {

                        let returnValue = parseJSON["status"] as? String

                        if(returnValue != "error")
                        {
                         self.delegate?.userLoggedIn(data: userEmail! )
                         UserDefaults.set(UserDefaults.standard)(true, forKey: "isUserLoggedIn");
                         self.dismiss(animated: true, completion: nil)

                        } else {
                            // display an alert message
                            let userMessage = parseJSON["message"] as? String
                            let myAlert = UIAlertController(title: "Alert", message: userMessage, preferredStyle: UIAlertControllerStyle.alert);
                            let okAction = UIAlertAction(title: "OK", style: UIAlertActionStyle.default, handler:nil)
                            myAlert.addAction(okAction);
                            self.present(myAlert, animated: true, completion: nil)
                        }

                    }
                } catch
                {
                    print(error)
                }

        }

    }

    task.resume()

}

Answer 1

你正在输出多个 json block ，这是非法的 json:

if(!empty($userDetails))
   ...
   echo json_encode(...)
} else {
   ...
   echo json_encode(...)
}
...
echo json_encode(...)

一个 JSON 文本 block 只能包含一个 SINGLE json 结构。因为你有两个，所以你有一个 JSON 语法错误。

例如

echo json_encode('hi');
echo json_encode('mom');

产生

"hi""mom"

而且由于 JSON 是 javascript 代码，您基本上是在尝试这样做:

var foo = "hi""mom";
              ^--syntax error, unexpected string