Problem encoding JSON for a UrlRequest in Swift

I'm trying to set up an API service. I have a simple login call that looks like this:

enum HttpMethod: Equatable {
case get([URLQueryItem])
case put(Data?)
case post(Data?)
var name: String {
switch self {
case .get: return "GET"
case .put: return "PUT"
case .post: return "POST"
}
}
}
struct Request<Response> {
let url: URL
let method: HttpMethod
var headers: [String: String] = [:]
}
extension Request {
static func postUserLogin(email: String, password:String) -> Self {
var params = [String: Any]()
params["username"] = email
params["password"] = password
let data = params.jsonData
return Request(
url: URL(string: NetworkingConstants.USER_LOGIN)!,
method: .post(
try? JSONEncoder().encode(data) //should this be something else???
)
)
}
}
extension Dictionary {
var jsonData: Data? {
return try? JSONSerialization.data(withJSONObject: self, options: [.prettyPrinted])
}
}
extension Request {
var urlRequest: URLRequest {
var request = URLRequest(url: url)
switch method {
case .post(let data), .put(let data):
request.httpBody = data
case let .get(queryItems):
//removed for brevity
default:
break
}
request.setValue("application/json", forHTTPHeaderField: "Content-Type")
request.allHTTPHeaderFields = headers
request.httpMethod = method.name
return request
}
}
extension URLSession {
func decode<Value: Decodable>(
_ request: Request<Value>,
using decoder: JSONDecoder = .init()
) async throws -> Value {
let decoded = Task.detached(priority: .userInitiated) {
let (data, _) = try await self.data(for: request.urlRequest)
try Task.checkCancellation()
return try decoder.decode(Value.self, from: data)
}
return try await decoded.value
}
}

The problem is, instead of the Body of the request being sent to the server looking like this:

{
"username": "me@you.com",
"password": "abcdef"
}

It looks like this:

"eyJwYXNzd29yZCI6ImFhYWEiLCJ1cZSI6InRhQHQuY29tIn0="

What am I doing wrong? Here's how I make the call:

let request: Request<UserLoginResponse> = .postUserLogin(email: "me@you.com", password: "abcdef")
let response = try? await URLSession.shared.decode(request)

request.httpBody looks like this:

▿ Optional<Data>
▿ some : 74 bytes
- count : 74
▿ pointer : 0x00007fcc74011200
- pointerValue : 140516096283136

Scott Thompson's answer is correct in the general case, and worth learning. In this specific case you can make it a bit simpler because you happen to have a [String: String] dictionary, which is already Encodable. So you don't need an extra encoding type:

static func postUserLogin(email: String, password:String) -> Self {
var params = [String: String]()
params["username"] = email
params["password"] = password
return Request(
url: URL(string: NetworkingConstants.USER_LOGIN)!,
method: .post(
try? JSONEncoder().encode(params)
)
)
}

Your problem is that you're first JSONSerialization-encoding params into a Data that contains the JSON characters. You then JSONEncoder-encoding that Data, which by default Base64-encodes it. That's the "ey..." output you're getting. There's no need for two steps. You can get rid of jsonData entirely, and just use JSONEncoder directly.

JSONEncoder and JSONDecoder are designed to work directly with Swift types allowing you to skip the need to build generic dictionaries or arrays as you are doing where you have this code:

// You should use a Swift type instead of this
var params = [String: Any]()
params["username"] = email
params["password"] = password
let data = params.jsonData

In your case you should declare a struct and mark it as Codable then encode that (from a playground):

struct UserAuthentication: Codable {
let username: String
let password: String
}
let email = "fred"
let password = "foobar"
let userAuthentication = UserAuthentication(username: email, password: password)
let encoder = JSONEncoder()
do {
let encoded = try encoder.encode(userAuthentication)
print(String(data: encoded, encoding: .utf8)!)
} catch (let error) {
print("Json Encoding Error \(error)")
}

Swift will represent your struct (in this case the UserAuthentication struct) as JSON using the names of the properties as the JSON keys and their values as the JSON values.

If you want to handle more complex cases, see:

Encoding and Decoding Custom Types