在制作自定义 iOS 键盘扩展时，如何创建一个与 Apple 原始退格键具有相同功能的按钮？

Question

我正在制作一个自定义键盘，它有用于删除文本的删除按钮，其代码是：

func addDelete() {

    deleteButton = UIButton.buttonWithType(.System) as UIButton
    deleteButton.setTitle(" Delete ", forState: .Normal)
    deleteButton.sizeToFit()
    deleteButton.setTranslatesAutoresizingMaskIntoConstraints(false)
    deleteButton.addTarget(self, action: "didTapDelete", forControlEvents: .TouchUpInside)
    deleteButton.layer.cornerRadius = 5

    view.addSubview(deleteButton)

    var rightSideConstraint = NSLayoutConstraint(item: deleteButton, attribute: .Right, relatedBy: .Equal, toItem: view, attribute: .Right, multiplier: 1.0, constant: -10.0)
    var topConstraint = NSLayoutConstraint(item: deleteButton, attribute: .Top, relatedBy: .Equal, toItem: view, attribute: .Top, multiplier: 1.0, constant: +10.0)
    view.addConstraints([rightSideConstraint, topConstraint])
}

func didTapDelete() {
    var proxy = textDocumentProxy as UITextDocumentProxy
    proxy.deleteBackward()
}

但这很慢，因为我必须点击很多次才能从 textField.

中删除很多字符

有什么方法可以创建一个我可以连续按下的按钮，它可以像默认的那样快速删除一个字符 iPhone keyBoard backSpace 按钮吗？

Answer 1

我从 HERE 中找到了答案。

在此键盘中，删除按钮具有与我想要的相同的功能。

该按钮的代码是：

private func addDeleteButton() {
    deleteButton = KeyButton(frame: CGRectMake(keyWidth * 8.5 + spacing * 9.5, keyHeight * 2.0 + spacing * 3.0 + predictiveTextBoxHeight, keyWidth * 1.5, keyHeight))
    deleteButton.setTitle("\u{0000232B}", forState: .Normal)
    deleteButton.addTarget(self, action: "deleteButtonPressed:", forControlEvents: .TouchUpInside)
    self.view.addSubview(deleteButton)

    let deleteButtonLongPressGestureRecognizer = UILongPressGestureRecognizer(target: self, action: "handleLongPressForDeleteButtonWithGestureRecognizer:")
    deleteButton.addGestureRecognizer(deleteButtonLongPressGestureRecognizer)

    let deleteButtonSwipeLeftGestureRecognizer = UISwipeGestureRecognizer(target: self, action: "handleSwipeLeftForDeleteButtonWithGestureRecognizer:")
    deleteButtonSwipeLeftGestureRecognizer.direction = .Left
    deleteButton.addGestureRecognizer(deleteButtonSwipeLeftGestureRecognizer)
}

func deleteButtonPressed(sender: KeyButton) {
    switch proxy.documentContextBeforeInput {
    case let s where s?.hasSuffix("    ") == true: // Cursor in front of tab, so delete tab.
        for i in 0..<4 { // TODO: Update to use tab setting.
            proxy.deleteBackward()
        }
    default:
        proxy.deleteBackward()
    }
    updateSuggestions()
}

func handleLongPressForDeleteButtonWithGestureRecognizer(gestureRecognizer: UILongPressGestureRecognizer) {
    switch gestureRecognizer.state {
    case .Began:
        if deleteButtonTimer == nil {
            deleteButtonTimer = NSTimer(timeInterval: 0.1, target: self, selector: "handleDeleteButtonTimerTick:", userInfo: nil, repeats: true)
            deleteButtonTimer!.tolerance = 0.01
            NSRunLoop.mainRunLoop().addTimer(deleteButtonTimer!, forMode: NSDefaultRunLoopMode)
        }
    default:
        deleteButtonTimer?.invalidate()
        deleteButtonTimer = nil
        updateSuggestions()
    }
}

func handleSwipeLeftForDeleteButtonWithGestureRecognizer(gestureRecognizer: UISwipeGestureRecognizer) {
    // TODO: Figure out an implementation that doesn't use bridgeToObjectiveC, in case of funny unicode characters.
    if let documentContextBeforeInput = proxy.documentContextBeforeInput as NSString? {
        if documentContextBeforeInput.length > 0 {
            var charactersToDelete = 0
            switch documentContextBeforeInput {
            case let s where NSCharacterSet.letterCharacterSet().characterIsMember(s.characterAtIndex(s.length - 1)): // Cursor in front of letter, so delete up to first non-letter character.
                let range = documentContextBeforeInput.rangeOfCharacterFromSet(NSCharacterSet.letterCharacterSet().invertedSet, options: .BackwardsSearch)
                if range.location != NSNotFound {
                    charactersToDelete = documentContextBeforeInput.length - range.location - 1
                } else {
                    charactersToDelete = documentContextBeforeInput.length
                }
            case let s where s.hasSuffix(" "): // Cursor in front of whitespace, so delete up to first non-whitespace character.
                let range = documentContextBeforeInput.rangeOfCharacterFromSet(NSCharacterSet.whitespaceCharacterSet().invertedSet, options: .BackwardsSearch)
                if range.location != NSNotFound {
                    charactersToDelete = documentContextBeforeInput.length - range.location - 1
                } else {
                    charactersToDelete = documentContextBeforeInput.length
                }
            default: // Just delete last character.
                charactersToDelete = 1
            }

            for i in 0..<charactersToDelete {
                proxy.deleteBackward()
            }
        }
    }
    updateSuggestions()
}

在制作自定义 iOS 键盘扩展时，如何创建一个与 Apple 原始退格键具有相同功能的按钮？

When making a custom iOS keyboard extension, how can I create a button that has the same functionality as Apple's original backspace key?

backspace

swift

custom-keyboard