Question

我是Swift 3的新手，我遇到了让Json返回然后发送请求的问题。我正在尝试使用参数用户名和密码向服务器发送一个帖子请求，并使用Json获取有关信息的响应，但我还没有能够返回数据。

输出：

错误：NSURLErrorDomain：-1003 状态码：200，标题连接=关闭; ＆＃34;内容类型＆＃34; =＆＃34; application / json＆＃34 ;; ＆＃34;传送编码＆＃34; =身份

Android上的

请求看起来像：

            "{\n" +
            "    \"jsonrpc\": \"2.0\",\n" +
            "    \"id\": \"1\",\n" +
            "    \"method\": \"call\",\n" +
            "    \"params\": [\n" +
            "        \"" + LOGIN_TOKEN + "\",\n" +
            "        \"session\",\n" +
            "        \"login\",\n" +
            "        {\n" +
            "            \"username\": \"" + userName + "\",\n" +
            "            \"password\": \"" + password + "\"\n" +
            "        }\n" +
            "    ]\n" +
            "}";

这是我要发送的请求：

"{ \"jsonrpc\": \"2.0\", \"id\": 1, \"method\": \"call\", \"params\": [ \"00000000000000000000000000000000\", \"session\", \"login\", { \"username\": \"root\", \"password\": \"admin01\"  } ] }"

这是我的代码：

   override func viewDidLoad() {
   var userName = "root"
   var password = "admin01"
   //var LOGIN_TOKEN = 0000000000000000

    let jsonObject: [String: Any] =
        ["jsonrpc" : 2.0,
         "id": 1,
         "method": "call",
         "params": [ "00000000000000",
                     "session",
                     "login",
                     [ "username": userName,
                       "password": password]],
         ]

    do {
        let jsonData = try JSONSerialization.data(withJSONObject: jsonObject, options: .prettyPrinted)
        // here "jsonData" is the dictionary encoded in JSON data

        let decoded = try JSONSerialization.jsonObject(with: jsonData, options: [])
        // here "decoded" is of type `Any`, decoded from JSON data

        // you can now cast it with the right type
        if let dictFromJSON = decoded as? [String:String] {
            // use dictFromJSON
        }
    } catch {
        print(error.localizedDescription)
    }

    Alamofire.request("http://192.168.1.1/ubus", method: .post, parameters: jsonObject, encoding: JSONEncoding.default)
        .responseJSON { response in
            print(response)
            //to get status code
            if let status = response.response?.statusCode {
                switch(status){
                case 201:
                    print("example success")
                default:
                    print("error with response status: \(status)")
                }
            }
            //to get JSON return value
            if let result = response.result.value {
                let JSON = result as! NSDictionary
                print(JSON)
            }

    }

    super.viewDidLoad()
    // Do any additional setup after loading the view, typically from a nib.
}

override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
    // Dispose of any resources that can be recreated.
}

Answer 1

您的JSON对象似乎无效

真正有用的JSON验证器是免费的。我想到的是http://jsonlint.com

我认为，根据您发布的代码我可以解释的是，您希望jsonObject看起来像这样，这是一个有效的JSON：

[
    {
    "jsonrpc": 2.0,
    "id": 1,
    "method": "call",
    "params": ["00000000000000", "session", "login"],
    "username": username,
    "password": password
    }
]

我认为username和password变量是您已分配的字符串？

Json解析Swift 3 Alamofire

1 个答案: