ios - 如何使 UIKeyboardType 仅用于十六进制输入？
<p><p>是否有任何选项可以使用扩展名或任何其他方式快速创建仅具有 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</p></p>
<br><hr><h1><strong>Best Answer-推荐答案</ strong></h1><br>
<p><p>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。</p>
<p>键盘是这样设计的:</p>
<p> <a href="/image/KaM7n.png" rel="noreferrer noopener nofollow"><img src="/image/KaM7n.png" alt="enter image description here"/></a> </p>
<p><em>设计基于 David Mulder 的布局</em>
<a href="https://ux.stackexchange.com/a/58605/128044" rel="noreferrer noopener nofollow">https://ux.stackexchange.com/a/58605/128044</a> </p>
<p>HexadecimalKeyboard 类创建键盘。</p>
<pre><code>protocol RemoveKeyboardDelegate: class {
func removeKeyboard()
}
class HexButton: UIButton {
var hexCharacter: String = ""
}
class HexadecimalKeyboard: UIView {
weak var target : UIKeyInput?
weak var delegate : RemoveKeyboardDelegate?
var hexadecimalButtons: = ["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)
}
}
//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)
}
}
//MARK: - Nest stackviews
mainStack.addArrangedSubview(panel1)
panel1.addArrangedSubview(hexadecimalButtons)
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
}
}
</code></pre>
<p><em>代码来源于 Rob 提供的十进制键盘示例</em> <a href="https://stackoverflow.com/a/57275689/1816667" rel="noreferrer noopener nofollow">https://stackoverflow.com/a/57275689/1816667</a> </p>
<p>下面是如何使用键盘的示例。在示例中，使用十六进制键盘设置了两个文本字段:</p>
<pre><code>class ViewController: UIViewController {
@IBOutlet var hexField: !
override func viewDidLoad() {
hexField.inputView= HexadecimalKeyboard(target: hexField)
hexField.inputView= HexadecimalKeyboard(target: hexField)
}
@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() }
}
}
</code></pre>
<p>此处提供了一个使用 Swift 5 的示例 Xcode 12 项目:
<a href="https://github.com/PepperoniJoe/HexadecimalKeyboard" rel="noreferrer noopener nofollow">https://github.com/PepperoniJoe/HexadecimalKeyboard</a> </p>
<p> <a href="/image/NPJuS.gif" rel="noreferrer noopener nofollow"><img src="/image/NPJuS.gif" alt="enter image description here"/></a> </p></p>
<p style="font-size: 20px;">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：
<a href="https://stackoverflow.com/questions/46579414/" rel="noreferrer noopener nofollow" style="color: red;">
https://stackoverflow.com/questions/46579414/
</a>
</p>
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