Question

swift中是否有任何选项可以使用扩展名或任何其他方式使用HEX（0-9，A，B，C，D，E，F）值创建新的UIKeyboardType？我想要只有十六进制字符启用的键盘，用户可以清楚地看到他只能输入十六进制字符，或者只能在键盘上看到十六进制字符

Answer 1

第一种情况，有如下所述的键盘类型，我们可以设置textField.keyboardType属性并使用它。

public enum UIKeyboardType : Int {    
    case `default` // Default type for the current input method.    
    case asciiCapable // Displays a keyboard which can enter ASCII characters    
    case numbersAndPunctuation // Numbers and assorted punctuation.    
    case URL // A type optimized for URL entry (shows . / .com prominently).    
    case numberPad // A number pad with locale-appropriate digits (0-9, ۰-۹, ०-९, etc.). Suitable for PIN entry.    
    case phonePad // A phone pad (1-9, *, 0, #, with letters under the numbers).    
    case namePhonePad // A type optimized for entering a person's name or phone number.
    case emailAddress // A type optimized for multiple email address entry (shows space @ . prominently).
    @available(iOS 4.1, *)
    case decimalPad // A number pad with a decimal point.
    @available(iOS 5.0, *)
    case twitter // A type optimized for twitter text entry (easy access to @ #)
    @available(iOS 7.0, *)
    case webSearch // A default keyboard type with URL-oriented addition (shows space . prominently).
    @available(iOS 10.0, *)
    case asciiCapableNumberPad // A number pad (0-9) that will always be ASCII digits.
    public static var alphabet: UIKeyboardType { get } // Deprecated
}

第二种情况是，您希望限制用户不要输入不需要的值（例如您只需要1-9和az），在这种情况下，您可以使用func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool委托和验证用户输入和输入字符是根据您的要求返回true否则返回false。

第三种情况如果使用给定键盘作为用户可选而不是更好地创建自定义键盘，请参阅此Apple developer reference。和nice tutorial我过去常常学习。

Answer 2

下面的代码创建了一个十六进制键盘，并将其传递给Saurabh Prajapati建议的inputView。

键盘的设计如下：

设计基于David Mulder的布局 https://ux.stackexchange.com/a/58605/128044

HexadecimalKeyboard类创建键盘。

protocol RemoveKeyboardDelegate: class {
    func removeKeyboard()
}

class HexButton: UIButton {
    var hexCharacter: String = ""
}

class HexadecimalKeyboard: UIView {
    weak var target   : UIKeyInput?
    weak var delegate : RemoveKeyboardDelegate?
    
    var hexadecimalButtons: [HexButton] = ["0","7","8","9","4","5","6","1","2","3","A","B","C","D","E","F"].map {
        let button = HexButton(type: .system)
        button.hexCharacter = $0
        button.setTitle("\($0)", for: .normal)
        button.backgroundColor = UIColor.secondarySystemGroupedBackground
        button.addTarget(self, action: #selector(didTapHexButton(_:)), for: .touchUpInside)
        return button
    }
    
    var deleteButton: UIButton = {
        let button = UIButton(type: .system)
        button.setTitle("⌫", for: .normal)
        button.backgroundColor = UIColor.systemGray4
        button.accessibilityLabel = "Delete"
        button.addTarget(self, action: #selector(didTapDeleteButton(_:)), for: .touchUpInside)
        return button
    }()
    
    var okButton: UIButton = {
        let button = UIButton(type: .system)
        button.setTitle("OK", for: .normal)
        button.backgroundColor = UIColor.systemGray4
        button.accessibilityLabel = "OK"
        button.addTarget(self, action: #selector(didTapOKButton(_:)), for: .touchUpInside)
        return button
    }()
    
    var mainStack: UIStackView = {
        let stackView = UIStackView()
        stackView.distribution = .fillEqually
        stackView.spacing      = 10
        stackView.autoresizingMask = [.flexibleWidth, .flexibleHeight]
        stackView.isLayoutMarginsRelativeArrangement = true
        stackView.layoutMargins = UIEdgeInsets(top: 10, left: 10, bottom: 10, right: 10)
        return stackView
    }()
    
    init(target: UIKeyInput) {
        self.target = target
        super.init(frame: .zero)
        configure()
    }
    
    required init?(coder: NSCoder) {
        fatalError("init(coder:) has not been implemented")
    }
}


// MARK: - Actions

extension HexadecimalKeyboard {
    @objc func didTapHexButton(_ sender: HexButton) {
        target?.insertText("\(sender.hexCharacter)")
    }
    
    @objc func didTapDeleteButton(_ sender: HexButton) {
        target?.deleteBackward()
    }
    
    @objc func didTapOKButton(_ sender: HexButton) {
        delegate?.removeKeyboard()
    }
}


// MARK: - Private initial configuration methods

private extension HexadecimalKeyboard {
    func configure() {
        self.backgroundColor = .systemGroupedBackground
        autoresizingMask     = [.flexibleWidth, .flexibleHeight]
        buildKeyboard()
    }
    
    func buildKeyboard() {
        //MARK: - Add main stackview to keyboard
        mainStack.frame = bounds
        addSubview(mainStack)
        
        //MARK: - Create stackviews
        let panel1         = createStackView(axis: .vertical)
        let panel2         = createStackView(axis: .vertical)
        let panel2Group    = createStackView(axis: .vertical)
        let panel2Controls = createStackView(axis: .horizontal, distribution : .fillProportionally)
 
        
        //MARK: - Create multiple stackviews for numbers
        for row in 0 ..< 3 {
            let panel1Numbers = createStackView(axis: .horizontal)
            panel1.addArrangedSubview(panel1Numbers)
            
            for column in 0 ..< 3 {
                panel1Numbers.addArrangedSubview(hexadecimalButtons[row * 3 + column + 1])
            }
        }
        
        //MARK: - Create multiple stackviews for letters
        for row in 0 ..< 2 {
            let panel2Letters = createStackView(axis: .horizontal)
            panel2Group.addArrangedSubview(panel2Letters)
            
            for column in 0 ..< 3 {
                panel2Letters.addArrangedSubview(hexadecimalButtons[9 + row * 3 + column + 1])
            }
        }
        
        //MARK: - Nest stackviews
        mainStack.addArrangedSubview(panel1)
        panel1.addArrangedSubview(hexadecimalButtons[0])
        mainStack.addArrangedSubview(panel2)
        panel2.addArrangedSubview(panel2Group)
        panel2.addArrangedSubview(panel2Controls)
        panel2Controls.addArrangedSubview(deleteButton)
        panel2Controls.addArrangedSubview(okButton)
        
        //MARK: - Constraint - sets okButton width to two times the width of the deleteButton plus 10 points for the space
        panel2Controls.addConstraint(NSLayoutConstraint(
                                        item       : okButton,
                                        attribute  : .width,
                                        relatedBy  : .equal,
                                        toItem     : deleteButton,
                                        attribute  : .width,
                                        multiplier : 2,
                                        constant   : 10))
    }
    
    func createStackView(axis: NSLayoutConstraint.Axis, distribution: UIStackView.Distribution = .fillEqually) -> UIStackView {
        let stackView = UIStackView()
        stackView.axis         = axis
        stackView.distribution = distribution
        stackView.spacing      = 10
        return stackView
    }
}

代码是从Rob https://stackoverflow.com/a/57275689/1816667

提供的十进制键盘示例中衍生而来的

然后是一个如何使用键盘的示例。在此示例中，使用十六进制键盘设置了两个文本字段：

class ViewController: UIViewController {

    @IBOutlet var hexField: [UITextField]!
    
    override func viewDidLoad() {
        hexField[0].inputView  = HexadecimalKeyboard(target: hexField[0])
        hexField[1].inputView  = HexadecimalKeyboard(target: hexField[1])
    }
    
    @IBAction func clickTextField(_ sender: UITextField) {
        sender.reloadInputViews()
        sender.inputView  = HexadecimalKeyboard(target: sender)
        
        let hexadecimalKeyboard  = HexadecimalKeyboard(target: sender)
                sender.inputView = hexadecimalKeyboard
                hexadecimalKeyboard.delegate = self
    }
} // end of View Controller

extension ViewController: RemoveKeyboardDelegate {
    func removeKeyboard() {
        _ = hexField.map { $0.inputView?.removeFromSuperview() }
    }
}

此处提供了使用Swift 5的示例Xcode 12项目： https://github.com/PepperoniJoe/HexadecimalKeyboard

如何仅为十六进制输入制作UIKeyboardType？

2 个答案: