如何使用 RxSwift 检测双击

Question

我正在尝试使用 RxSwift 检测双击

如果没有 RxSwift，我会这样：

private func setupFakePanView() {
    let singleTapGesture = UITapGestureRecognizer()
    let doubleTapGesture = UITapGestureRecognizer()

    singleTapGesture.numberOfTapsRequired = 1
    doubleTapGesture.numberOfTapsRequired = 2

    singleTapGesture.addTarget(self, action: #selector(self.tapped))
    doubleTapGesture.addTarget(self, action: #selector(self.doubleTapped))

    someView.addGestureRecognizer(singleTapGesture)
    someView.addGestureRecognizer(doubleTapGesture)

    singleTapGesture.require(toFail: doubleTapGesture)
}

@objc private func tapped() {
    // Do something
}

@objc private func doubleTapped() {
    // Do something else
}

有没有办法用 RxSwift、RxCocoa 和RxGesture实现同样的效果？ 我已经尝试了以下但当然它不起作用：

someView.rx
    .tapGesture(numberOfTouchesRequired: 1, numberOfTapsRequired: 1)
    .when(.recognized)
    .subscribe(onNext: { _ in
        // Do something
    })
    .disposed(by: bag)

someView.rx
    .tapGesture(numberOfTouchesRequired: 1, numberOfTapsRequired: 2)
    .when(.recognized)
    .subscribe(onNext: { _ in
        // Do something else
    })
    .disposed(by: bag)

有没有办法让第一个 tapGesture 知道第二个必须失败？

Answer 1

我找到了 2 个解决方案来解决这个问题！

A. 使用自定义UITapGestureRecognizer

let doubleTapGesture = UITapGestureRecognizer()
doubleTapGesture.numberOfTapsRequired = 2

let singleTapGesture = UITapGestureRecognizer()
singleTapGesture.numberOfTapsRequired = 1
singleTapGesture.require(toFail: doubleTapGesture)

let singleTap = someView.rx
    .gesture(singleTapGesture)
    .when(.recognized)
    .subscribe(onNext: { _ in
        // Do something
    })
    .disposed(by: bag)


let doubleTap = someView.rx
    .gesture(doubleTapGesture)
    .when(.recognized)
    .subscribe(onNext: { _ in
        // Do something else
    })
    .disposed(by: bag)

要么..

B. 使用自定义的UIGestureRecognizerDelegate

感谢Kishan建议Jegnux 的回答！

1 - 为单击手势设置自定义委托...

someView.rx
    .tapGesture(
        numberOfTouchesRequired: 1,
        numberOfTapsRequired: 1,
        configuration: { [weak self] gesture, delegate in
            gesture.delegate = self
        }
    )
    .subscribe(onNext: { _ in
        // Do something
    })
    .disposed(by: bag)

// double tap same as before

2 - 实施gestureRecognizer(_:shouldRequireFailureOf:)

extension MyController: UIGestureRecognizerDelegate {
    func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldRequireFailureOf otherGestureRecognizer: UIGestureRecognizer) -> Bool {
        if let gesture = otherGestureRecognizer as? UITapGestureRecognizer, gesture.numberOfTapsRequired == 2 {
             return true
        }
        return false
    }
}

两种解决方案都可以正常工作。

Answer 2

此代码应该工作：

tap
    .flatMapFirst {
        tap
        .takeUntil(tap.startWith(()).debounce(.milliseconds(300), scheduler: MainScheduler.instance))
        .startWith(())
        .reduce(0) { acc, _ in acc + 1 }
    }
    .map { min($0, 2) }

只需使用一个手势识别器（其事件称为tap ），它会在点击发生时立即发出。 map上方的代码将此问题概括为 output 在这种情况下，彼此在特定时间段（300 毫秒）内连续点击的次数。 map只是为了保证只有一个1或者一个2出来。 然后随心所欲地执行任何条件逻辑。 我成功地用UIButton测试了它。