Construction was too complex to be solved in reasonable time - ios

Question

I have problem with Swift 3, I am trying to send a request to a server and get JSON, but I get:
Construction was too complex to be solved in reasonable time.
I tried every way, but it doesn't worked.
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)")
}
}
Now I tryed to creat dic and convert to Json, but after that, I get problem on request there I declare my parameters. they say: use of unresolved identifier dictFromJSON
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)")

Answer 1

UPDATED:
According to the document of Alamofire, (you can see here), you don't need to convert the parameters Dictionary to JSON.
For example,
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
OLD:
You should use Dictionay for the parameter actually.
Therefore, instead of declaring the parameter like you did, you should do like this:
let parameters = ["jsonrpc" : 2.0,
"id": 1,
"method": "call",
"params": [ "00000000000000",
"session",
"login",
[ "username": userName,
"password": password]],
]