Dan*_*bek 1 xcode login parse-platform swift
我在xcode上使用swift和解析,我一直收到这个错误:
[Error]: invalid login parameters (Code: 101, Version: 1.7.2)
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每当我尝试登录用户时,我知道用户是在解析后端创建的,并且其信息是正确的.我该怎么做才能阻止这种情况发生并让用户在没有应用程序发回无效登录参数的情况下登录?
这是我的LogInViewController:
import Foundation
import Parse
import UIKit
import Bolts
class LoginViewController: UIViewController, UITextFieldDelegate {
@IBOutlet weak var loginStatusLabel: UILabel!
@IBOutlet weak var emailTextField: UITextField!
@IBOutlet weak var passwordTextField: UITextField!
@IBOutlet weak var loginButton: UIButton!
override func viewDidLoad() {
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.
emailTextField.delegate = self
passwordTextField.delegate = self
}
override func didReceiveMemoryWarning() {
super.didReceiveMemoryWarning()
// Dispose of any resources that can be recreated.
}
@IBAction func loginButtonPress(sender: AnyObject) {
login(emailTextField.text, password: passwordTextField.text)
}
func login(email: String, password: String)
{
PFUser.logInWithUsernameInBackground("email", password: "password")
{
(user: PFUser?, error: NSError?) -> Void in
if user != nil
{
user!.fetch()
if user!.objectForKey("emailVerified") as! Bool
{
self.loginStatusLabel.text = "Success!"
}
else if !(user!.objectForKey("emailVerified") as! Bool)
{
self.loginStatusLabel.text = "Verify your email address!"
}
else // status is "missing"
{
//TODO: Handle this error better
self.loginStatusLabel.text = "Verification status: Missing"
}
}
else
{
if let errorField = error!.userInfo
{
self.loginStatusLabel.text = (errorField["error"] as! NSString) as String
}
else
{
// No userInfo dictionary present
// Help from http://stackoverflow.com/questions/25381338/nsobject-anyobject-does-not-have-a-member-named-subscript-error-in-xcode
}
}
}
}
func textFieldShouldReturn(textField: UITextField) -> Bool {
textField.resignFirstResponder()
return true;
}
}
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我在登录用户时出错了?
您通过串"email"并"password"分析和不变量的实际值email和password.
将您的代码更改为:
PFUser.logInWithUsernameInBackground(email, password: password)
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