Ios 施工过于复杂，无法在合理时间内解决

我对Swift 3有问题，我试图向服务器发送请求并获取JSON，但我得到：

施工过于复杂，无法在合理的时间内解决

我试过各种方法，但都不管用

var userName = "root"
var password = "admin01"
//var LOGIN_TOKEN = 0000000000000000

let parameters = [
    "{\n",
    "    \"jsonrpc\": \"2.0\",\n",
    "    \"id\": \"1\",\n",
    "    \"method\": \"call\",\n",
    "    \"params\": [\n",
    "        \"0000000000000000\",\n",
    "        \"session\",\n",
    "        \"login\",\n",
    "        {\n",
    "            \"username\": \"" + userName + "\",\n",
    "            \"password\": \"" + password + "\"\n",
    "        }\n",
    "    ]\n",
    "}"
]

let joiner = ""
let joinedStrings = parameters.joined(separator: joiner)
print("joinedStrings: \(joinedStrings)")

// All three of these calls are equivalent
Alamofire.request("http://192.168.1.1", method: .post, parameters: parameters).responseJSON { response in
    print("Request: \(response.request)")
    print("Response: \(response.response)")


    if let JSON = response.result.value {
        print("JSON: \(JSON)")
    }
}

现在我尝试创建dic并转换为Json，但在那之后，我在请求时遇到了问题，我声明了我的参数。他们说：使用未解析的标识符

  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)
    }




    // All three of these calls are equivalent
    Alamofire.request("http://192.168.1.1/ubus", method: .post, parameters: dictFromJSON).responseJSON { response in
        print("Request: \(response.request)")
        print("Response: \(response.response)")

更新：

根据Alamofire的文档（您可以看到），您不需要将参数字典转换为JSON。

比如说,

let parameters: Parameters = [
    "foo": "bar",
    "baz": ["a", 1],
    "qux": [
        "x": 1,
        "y": 2,
        "z": 3
    ]
]

// All three of these calls are equivalent
Alamofire.request("https://httpbin.org/post", parameters: parameters)
Alamofire.request("https://httpbin.org/post", parameters: parameters, encoding: URLEncoding.default)
Alamofire.request("https://httpbin.org/post", parameters: parameters, encoding: URLEncoding.httpBody)

// HTTP body: foo=bar&baz[]=a&baz[]=1&qux[x]=1&qux[y]=2&qux[z]=3

旧的：

实际上，您应该为参数使用术语

因此，您不应该像以前那样声明参数，而是应该这样做：

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

---

无论如何，不要手动创建JSON字符串。创建数组和字典，然后将它们转换为JSON。现在我得到一个错误：调用中有额外的参数'method'。你能帮我吗？这只是Swift 2中的一个例子。重要的是想法。你可以自己找到其他的例子，有很多。我创建了dic并转换为JSon，但我有另一个问题，请检查我的问题，我更新了它。我尝试过这种方式，但现在我的请求有问题，我更新了我的问题，请检查，并帮助我