我我的类用户地图与对象映射:无法转换'String?'类型的值(Alamofire)
class User: Mappable{
private var _username: String! = nil
private var _password: String! = nil
private var _firstName: String! = nil
private var _lastName: String! = nil
//here are getters and setters(it takes too much space)
init(){
}
required init?(_ map: Map) {
}
func mapping(map: Map) {
username <- map["USERNAME"]
password <- map["PASSWORD"]
firstName <- map["FIRST_NAME"]
lastName <- map["LAST_NAME"]
}
}
然后我试图让新用户,并把一些值,映射整个对象,并使用Alamofire发送这样的:
let userEmail = userEmailField.text!
let userPassword = userPasswordField.text!
let user = User()
user.username = userEmail
user.password = userPassword
let JSONString = Mapper().toJSONString(user, prettyPrint: true)
AlamofireService.alamofireService.makePostServiceRequest(URL_BASE, parameters: JSONString, resposeCallback: self)
我遵循库的指示,但我得到错误“无法转换类型'字符串的值?'到期望的参数类型'[String:AnyObject]'“,为什么?
什么是您会收到错误? –
@MatanLachmish对不起,我忘了写。我编辑我的问题。 –
使用私有实例变量与getter和setters是什么意思?这是Swift! init方法中的隐式unwrapped option是什么意思?这两个值都可以是“nil”,然后将它们声明为** real **可选,或者如果它们从不'nil'则将它们声明为非可选。 – vadian