制作类似于锁定屏幕的登录屏幕

Question

我正在尝试使用 4 个文本字段制作类似于锁定屏幕的登录屏幕。我面临的问题是将焦点从一个文本字段更改为下一个文本字段。

当我说 secondTextField.becomeFirstResponder() 时，firstTextField 的值被复制到 secondTextField
我正在使用小数点键盘类型

这是我到目前为止所做的，但我找不到解决问题的方法：

使用 UITextFieldDelegate
为所有文本字段建立了 IBOutlet 连接
firstText.delegate = self secondText.delegate = self thirdText.delegate = self firstText.becomeFirstResponder()

func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool { var newLength = countElements(textField.text) + countElements(字符串) - range.length

    if newLength == 1 {
        if textField == firstText{
            changeFocus(textField)
        }else if textField == secondText {
            changeFocus(textField)
        }else if textField == thirdText {
            changeFocus(textField)
        }

    }
return newLength <= 1

}

func changeFocus(nextFocus: UITextField){

if nextFocus == firstText {
    firstText.resignFirstResponder()
    secondText.becomeFirstResponder()
} else if nextFocus == secondText {
    secondText.resignFirstResponder()
    thirdText.becomeFirstResponder()
} else if nextFocus == thirdText {
    thirdText.resignFirstResponder()
}

}

我怎样才能正确地做到这一点，请提供任何帮助和建议。

Answer 1

使用此委托方法更改焦点

func textFieldShouldReturn(textField: UITextField) -> Bool

Answer 2

要向 svrushal 添加一点答案：

func textFieldShouldReturn(textField: UITextField) -> Bool {
   if textField == firstText {
       secondText.becomeFirstResponder()
   }

   if textField == secondText {
       thirdText.becomeFirstResponder()
   }

   if textField == thirdText {
       fourthText.becomeFirstResponder()
   }

}

Answer 3

你可以参考这个linkhttps://github.com/dhennessy/PAPasscode/blob/master/PAPasscode/PAPasscodeViewController.h

它会帮助你。

Answer 4

我在这里找到了类似的东西：similar

成功运行，但有两个限制：

首先，我无法向后删除
其次，如果密码错误，用户必须点击第一个文本字段。但现在这些已经足够好了。

方法如下

func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {

var shouldProcess = false
var moveToNextTextField = false
var insertStringLength = countElements(string)
if insertStringLength == 0 {
    shouldProcess = true
}
else {
    if countElements(textField.text) == 0 {
        shouldProcess = true
    }
}
if shouldProcess {
   var myString = textField.text
    if countElements(myString) == 0{
        myString = myString + string

        password = password + myString
        moveToNextTextField = true
    }
    textField.text = myString

    if moveToNextTextField {
        changeFocus(textField)
    }
}
return false
}

改变焦点的方法：

func changeFocus(nextTextField : UITextField){
if  nextTextField == firstText {
    firstText.resignFirstResponder()
    secondText.becomeFirstResponder()
} else if nextTextField == secondText {
    secondText.resignFirstResponder()
    thirdText.becomeFirstResponder()
} else if nextTextField == thirdText {
    thirdText.resignFirstResponder()
    fourthText.becomeFirstResponder()
} else if nextTextField == fourthText {
    fourthText.resignFirstResponder()
}

}

检查方法 pin/passcode :

func textFieldShouldEndEditing(textField: UITextField) -> Bool {
if textField == fourthText {
   thirdText.resignFirstResponder()
    let pass = defaultUser.stringForKey(Admin)
    println("password = \(password)")
    if pass  == password {

        println("the password is correct")
        dispatch_async(dispatch_get_main_queue()) { () -> Void in
            self.performSegueWithIdentifier("SegueToSecond", sender: self)
        }

    } else {
        println("wrong password")
        thirdText.text = ""
        secondText.text = ""
        firstText.text = ""
        fourthText.text = ""
        password = ""
        view.reloadInputViews()
    }

}

return true
}

希望这些对以后的人有所帮助。

制作类似于锁定屏幕的登录屏幕

Making a Login screen similar to the lock screen

ios

swift

uitextfield