Question

我想在请求中发送两个值：

1）操作字符串

2）用户对象

即使我在参数Operation is not set

中加入了"operation": "register"，我也收到了消息.Success

我是Alamofire的新手。任何人都可以向我解释：

1）如何在请求中发送值？

2）如何发送用户对象？

3）如何处理结果.Failure和let urlString = URLFactory() let url = URL(string: urlString.getAppURL())! print("Log url: \(url)") let user = User() user.setEmail(email: email) let parameters: Parameters = ["operation": "register", "user": user] Alamofire.request(url, method: .post, parameters: parameters).responseJSON { response in print("Log \(response)") print("Log response.request: \(response.request)") print("Log response.error: \(response.error)") print("Log response.data: \(response.data)") print("Log response.result: \(response.result)") }

Swift Code：

Log url: http://192.168.0.101/GeolocationNews/NewsDataCrawler/app.php
Log SUCCESS: {
    message = "Invalid Parameters";
    result = failure;
}
Log response.request: Optional(http://192.168.0.101/GeolocationNews/NewsDataCrawler/app.php)
Log response.error: nil
Log response.data: Optional(51 bytes)
Log response.result: SUCCESS

快速输出：

$login = new Login();
$fun = new FunctionsValidation();

if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    $data = json_decode(file_get_contents("php://input"));

    if(isset($data->operation)) {
        $operation = $data->operation;
        if(!empty($operation)) {
            if($operation == 'register') {
                echo $login->register($data);
            }
        } else { // if operation is empty
            $response["result"] = "failure";
            $response["message"] = "Operation is empty";
            echo json_encode($response);
        }
    } else { // if operation is not set
        $response["result"] = "failure";
        $response["message"] = "Operation is not set";
        echo json_encode($response);
    }
}

PHP代码：

{
    "operation": "register",
    "user":
    {
        "email": "email value"
    }
}

更新

我已通过Postman发送测试API：

{"result":"failure","message":"Invalid Email"}

它给了我：class User: NSObject { private var name: String, email: String, password: String, oldPassword: String, newPassword: String, code: String private var id: Int override init() { self.name = "" self.email = "" self.password = "" self.oldPassword = "" self.newPassword = "" self.code = "" self.id = 0 } // set and get methods ... }所以API工作正常！

我尝试通过参数操作发送Alamofire请求并且它有效。所以似乎问题在于将用户对象转换为字典。谁能给我一个如何做到这一点的例子？

用户对象：

os.system()

Answer 1

我认为问题在于编码。根据您的PHP代码，它接受application/json作为内容类型，这应该通过Almofire使用JSON编码发送。

请改为尝试：

Alamofire.request(url, method: .post, parameters: parameters, encoding: JSONEncoding.default)
    .responseJSON { response in
        print("Log \(response)")
        print("Log response.request: \(response.request)")
        print("Log response.error: \(response.error)")
        print("Log response.data: \(response.data)")
        print("Log response.result: \(response.result)")
    }

<强>参考： https://github.com/Alamofire/Alamofire/blob/master/Documentation/Usage.md#parameter-encoding

Answer 2

问题在于将用户对象转换为字典。我只是将用户设置为字典而不是使用对象。

let userDictionary: Dictionary = ["email": email, "password": password]
let parameters: Parameters = ["operation": operation, "user": userDictionary]
Alamofire.request(url, method: .post, parameters: parameters, encoding: JSONEncoding.default).responseJSON { response in
    ...
}

Swift在Alamofire中发送参数

2 个答案: