php - 在 Alamofire 中快速发送参数

Question

我想在请求中发送两个值:

1) 操作字符串

2)用户对象

即使我在参数中包含了“操作”:“注册”，我仍收到消息操作未设置

我是阿拉莫菲尔的新手。谁能给我解释一下:

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.Success和.Failure

Swift 代码:

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)")
}

快速输出:

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

PHP 代码:

$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);
    }
}

更新

我已经通过Postman发送测试了API:

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

它给了我:{"result":"failure","message":"Invalid Email"} 所以 API 运行良好!

我尝试过仅通过参数中的操作发送 Alamofire 请求，并且它有效。所以看来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:

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 ...
}

Answer 1

我认为问题出在编码上。根据您的 PHP 代码，它接受 application/json 作为内容类型，并且应使用 JSON 编码通过 Almofire 发送。

试试这个:

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)")
    }

引用: https://github.com/Alamofire/Alamofire/blob/master/Documentation/Usage.md#parameter-encoding