Deserialize JSON / NSDictionary to Swift objects

SWIFT 4 Update

Since you give a very simple JSON object the code prepared for to handle that model. If you need more complicated JSON models you need to improve this sample.

Your Custom Object

class Person : NSObject {
    var name : String = ""
    var email : String = ""
    var password : String = ""

    init(JSONString: String) {
        super.init()

        var error : NSError?
        let JSONData = JSONString.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)

        let JSONDictionary: Dictionary = NSJSONSerialization.JSONObjectWithData(JSONData, options: nil, error: &error) as NSDictionary

        // Loop
        for (key, value) in JSONDictionary {
            let keyName = key as String
            let keyValue: String = value as String

            // If property exists
            if (self.respondsToSelector(NSSelectorFromString(keyName))) {
                self.setValue(keyValue, forKey: keyName)
            }
        }
        // Or you can do it with using 
        // self.setValuesForKeysWithDictionary(JSONDictionary)
        // instead of loop method above
    }
}

And this is how you invoke your custom class with JSON string.

override func viewDidLoad() {
    super.viewDidLoad()
    let jsonString = "{ \"name\":\"myUser\", \"email\":\"[email protected]\", \"password\":\"passwordHash\" }"
    var aPerson : Person = Person(JSONString: jsonString)
    println(aPerson.name) // Output is "myUser"
}

I recommend that you use code generation (http://www.json4swift.com) to create native models out of the json response, this will save your time of parsing by hand and reduce the risk of errors due to mistaken keys, all elements will be accessible by model properties, this will be purely native and the models will make more sense rather checking the keys.

Your conversion will be as simple as:

let userObject = UserClass(userDictionary)
print(userObject!.name)