快速编译错误: Downcast from 'String?!' to 'String' only unwraps optionals; did you mean to use '!!' ?

Question

升级到 cocoapods 1.0 后，我得到以下代码行的编译错误:

var strName = String()
var strEmail = String()
var strFacebookID = String()
var strPassword = String()
var objHelper = Helper()

....

let strFirstName = result["first_name"] as! String
let strLastName = result["last_name"] as! String
self.strName = strFirstName + "_" + strLastName
self.strEmail = result["email"] as! String
self.strFacebookID = result["id"] as! String

Downcast from 'String?!' to 'String' only unwraps optionals; did you mean to use '!!'?

这里是错误的详细截图: http://imgur.com/Efe1nQf

更新:更多代码在这里:https://gist.github.com/anonymous/9c91c2eb1ccf269e78a118970468d1e8

Answer 1

错误消息说 result 本身是可选的，因此您必须分别解包 result 和值。

let strFirstName = result!["first_name"] as! String

或更好地使用可选绑定(bind)以获得更高的安全性和更少的类型转换

if let userData = result as? [String:String] {
   let strFirstName = userData["first_name"]!
   let strLastName = userData["last_name"]!
}