我正在尝试创建一个简单的用户登录ViewController,它连接到充满我的应用程序用户信息的数据库,但是出现标题中指出的错误。
import UIKit
import Alamofire
class Users: Decodable {
let username: String
let email: String
let password: String
init(username: String, email: String, password: String) {
self.username = username
self.email = email
self.password = password
}
class LoginVC: UIViewController {
var loggingin = [Users]()
@IBOutlet weak var usernameTxtField: UITextField!
@IBOutlet weak var passwordTxtField: UITextField!
@IBOutlet weak var checkCredsBtn: UIButton!
override func viewDidLoad() {
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.
let jsonURL = "http://host-2:8888/getLogin.php"
let url = URL(string: jsonURL)
URLSession.shared.dataTask(with: url!) { (data, response, error) in
do {
self.loggingin = try JSONDecoder().decode([Users].self, from: data!)
for eachUser in self.loggingin {
print(eachUser.username + " : " + eachUser.password)
}
}
catch {
print("Error")
}
}.resume()
}
@IBAction func checkCreds(_ sender: Any) {
if usernameTxtField == loggingin.username && passwordTxtField == loggingin.password {
print("YES")
}
}
}
}
最佳答案
编译时间错误的原因出现在checkCreds
函数中,您试图直接从数组中访问username
和password
属性,这显然是不正确的。相反,您应该从loggingin
数组中获取所需的对象,并对其属性进行比较:
@IBAction func checkCreds(_ sender: Any) {
let currentUser = loggingin[0]
if usernameTxtField == currentUser.username && passwordTxtField == currentUser.password {
print("YES")
}
}
在上面的示例中,我刚得到第一个对象;我假设您已经能够获取所需的对象
currentUser
(使用的索引),以便从loggingin
获取对象。
关于swift - 类型 '[Users]'的值没有成员 'username',我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49466654/