UILabel
是否有任何可以设置的值以使其可选?
我有一个我想要选择的标签,(长按和复制btn显示)有点像Safari。
答案 0 :(得分:3)
是的,您需要从应用于UILabel的长按手势实现UIMenuController。在NSHipster上有一篇关于此的优秀文章,但文章的要点如下。
创建UILabel的子类并实现以下方法:
override func canBecomeFirstResponder() -> Bool {
return true
}
override func canPerformAction(action: Selector, withSender sender: AnyObject?) -> Bool {
return (action == "copy:")
}
// MARK: - UIResponderStandardEditActions
override func copy(sender: AnyObject?) {
UIPasteboard.generalPasteboard().string = text
}
然后在视图控制器中,您可以为标签添加长按手势:
let gestureRecognizer = UILongPressGestureRecognizer(target: self, action: "handleLongPressGesture:")
label.addGestureRecognizer(gestureRecognizer)
用这种方法处理长按:
func handleLongPressGesture(recognizer: UIGestureRecognizer) {
if let recognizerView = recognizer.view,
recognizerSuperView = recognizerView.superview
{
let menuController = UIMenuController.sharedMenuController()
menuController.setTargetRect(recognizerView.frame, inView: recognizerSuperView)
menuController.setMenuVisible(true, animated:true)
recognizerView.becomeFirstResponder()
}}
注意:此代码直接来自NSHipster文章,我只是将其包含在此处以符合SO标准。
答案 1 :(得分:2)
您可以改编@BJHSolutions和NSHipster的解决方案,以制作以下独立的SelectableLabel
:
import UIKit
/// Label that allows selection with long-press gesture, e.g. for copy-paste.
class SelectableLabel: UILabel {
override func awakeFromNib() {
super.awakeFromNib()
isUserInteractionEnabled = true
addGestureRecognizer(
UILongPressGestureRecognizer(
target: self,
action: #selector(handleLongPress(_:))
)
)
}
override var canBecomeFirstResponder: Bool {
return true
}
override func canPerformAction(_ action: Selector, withSender sender: Any?) -> Bool {
return action == #selector(copy(_:))
}
// MARK: - UIResponderStandardEditActions
override func copy(_ sender: Any?) {
UIPasteboard.general.string = text
}
// MARK: - Long-press Handler
@objc func handleLongPress(_ recognizer: UIGestureRecognizer) {
if recognizer.state == .began,
let recognizerView = recognizer.view,
let recognizerSuperview = recognizerView.superview {
UIMenuController.shared.setTargetRect(recognizerView.frame, in: recognizerSuperview)
UIMenuController.shared.setMenuVisible(true, animated:true)
recognizerView.becomeFirstResponder()
}
}
}
答案 2 :(得分:0)
UILabel继承自UIView,因此您只需向标签添加长按手势识别器即可。请注意,您必须将isUserInteractionEnabled更改为true,因为标签的默认值为false。
import UIKit
class ViewController: UIViewController {
let label = UILabel()
override func viewDidLoad() {
view.addSubview(label)
label.text = "hello"
label.translatesAutoresizingMaskIntoConstraints = false
label.centerXAnchor.constraint(equalTo: view.centerXAnchor).isActive = true
label.centerYAnchor.constraint(equalTo: view.centerYAnchor).isActive = true
let longPressGestureRecognizer = UILongPressGestureRecognizer(target: self, action: #selector(longPressLabel(longPressGestureRecognizer:)))
label.addGestureRecognizer(longPressGestureRecognizer)
label.isUserInteractionEnabled = true
}
@objc private func longPressLabel (longPressGestureRecognizer: UILongPressGestureRecognizer) {
if longPressGestureRecognizer.state == .began {
print("long press began")
} else if longPressGestureRecognizer.state == .ended {
print("long press ended")
}
}
}
答案 3 :(得分:-1)
我已经实现了一个UILabel
子类,它提供了所需的所有功能。请注意,如果您在界面构建器中使用它,则需要调整init
方法。
/// A label that can be copied.
class CopyableLabel: UILabel
{
// MARK: - Initialisation
/// Creates a new label.
init()
{
super.init(frame: .zero)
let gestureRecognizer = UILongPressGestureRecognizer(target: self, action: #selector(handleLongPressGesture(_:)))
self.addGestureRecognizer(gestureRecognizer)
self.isUserInteractionEnabled = true
}
required init?(coder aDecoder: NSCoder)
{
fatalError("init(coder:) has not been implemented")
}
// MARK: - Responder chain
override var canBecomeFirstResponder: Bool
{
return true
}
// MARK: - Actions
/// Method called when a long press is triggered.
func handleLongPressGesture(_ gestuerRecognizer: UILongPressGestureRecognizer)
{
guard let superview = self.superview else { return }
let menuController = UIMenuController.shared
menuController.setTargetRect(self.frame, in: superview)
menuController.setMenuVisible(true, animated:true)
self.becomeFirstResponder()
}
override func copy(_ sender: Any?)
{
UIPasteboard.general.string = self.text
}
}