swift 中是否有任何选项可以使用扩展名或任何其他方式制作仅具有 HEX(0-9,A,B,C,D,E,F) 值的新 UIKeyboardType？我想要只启用十六进制字符的键盘，用户可以清楚地看到他只能输入十六进制字符，或者键盘上只能看到十六进制字符

最佳答案

按照 Saurabh Prajapati 的建议，以下代码创建一个十六进制键盘并将其传递给 inputView。

键盘是这样设计的:

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
    }
}

代码源自 Rob 提供的十进制键盘示例 https://stackoverflow.com/a/57275689/1816667

下面是一个如何使用键盘的例子。在示例中，两个文本字段设置为十六进制键盘:

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() }
    }
}

此处提供了一个使用 Swift 5 的示例 Xcode 12 项目: https://github.com/PepperoniJoe/HexadecimalKeyboard

关于ios - 如何使 UIKeyboardType 仅用于十六进制输入？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/46579414/