我正在寻找在某些错误代码出现在我的程序中时创建某些事件。例如,如果有错误代码200,我需要让用户知道他们错过了用户名字段。或者对于错误代码125,我需要让他们知道在创建帐户时他们没有输入有效的电子邮件地址。我如何专门针对这些错误代码?我试过下面的代码没有成功,我做错了什么,这可能吗?在解析中处理某些错误
if error.code == 125 {
var invalidEmail:UIAlertView = UIAlertView(title: Please try again, message: "That does not look like a real email address. Please enter a real one.", delegate: self, cancelButtonTitle: "Try again")
invalidEmail.show()
}
xCode告诉我的错误是我有一个无法解析的标识符'错误'。在解析启动器项目文件'AppDelegate.swift'中有一个这样的实例,它调用以下代码,它似乎很好地工作。
func application(application: UIApplication, didFailToRegisterForRemoteNotificationsWithError error: NSError) {
if error.code == 3010 {
println("Push notifications are not supported in the iOS Simulator.")
} else {
println("application:didFailToRegisterForRemoteNotificationsWithError: %@", error)
}
}
我的代码
@IBAction func signupTapped(sender: AnyObject) {
let fullname = fullnameField.text
let email = emailField.text
let username = usernameField.text
let password = passwordField.text
var user = PFUser()
user.username = username
user.password = password
user.email = email
// other fields can be set just like with PFObject
user["fullname"] = fullname
if error.code == 125 {
let alert = UIAlertController(title: "Please try again", message: "That does not look like a valid email address.", preferedStyle: .Alert)
alert.addAction(UIAlertAction(title: "Cancel", style: .Cancel, handler: nil))
presentViewController(alert, animated: true)
}
if fullname.isEmpty || email.isEmpty || username.isEmpty || password.isEmpty {
let emptyFieldsError:UIAlertView = UIAlertView(title: "Please try again", message: "Please fill out all the fields so that we can create your account.", delegate: self, cancelButtonTitle: "Try again")
emptyFieldsError.show()
}
user.signUpInBackgroundWithBlock {
(succeeded: Bool, error: NSError?) -> Void in
if let error = error {
let errorString = error.userInfo?["error"] as? NSString
// Show the errorString somewhere and let the user try again.
} else {
// Hooray! Let them use the app now.
}
}
}
你忘了打电话给'show'消息吗? – tounaobun
你的'UIAlertView'似乎不会调用'show',而这正是不推荐使用的。 – Larme
我已根据收到的意见对我所做的实现进行了更改。 –