UITapGestureRecognizer点击self.view但忽略子视图

Question

当我双击self.view（UIViewCotroller的视图）时，我需要实现一个将调用某些代码的功能。但是我在这个视图上有其他UI对象的问题，我不想将任何识别器对象附加到所有这些对象上。我在下面找到了这个方法如何在我的视图上做手势，我知道它是如何工作的。现在我在障碍面前选择哪种方式来创建忽略子视图的识别器。有任何想法吗？谢谢。

UITapGestureRecognizer *doubleTap = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(handleDoubleTap:)];
[doubleTap setNumberOfTapsRequired:2];
[self.view addGestureRecognizer:doubleTap];

Answer 1

您应该在UIGestureRecognizerDelegate对象中采用self协议，并调用以下方法来检查视图。在此方法中，检查您对touch.view的视图并返回相应的bool（是/否）。像这样的东西：

- (BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch
{
    if ([touch.view isDescendantOfView:yourSubView]) {
        return NO;
    }
    return YES;
}

编辑：拜托，请检查@ Ian的答案！

Answer 2

斯威夫特4：

touch.view现在是可选的，所以基于@Antoine的回答：

func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
    if let touchedView = touch.view, touchedView.isDescendant(of: deductibleBackgroundView) {
        return false
    }
    return true
}

Answer 3

我不得不阻止儿童视图上的手势。唯一有效的方法是在所有下一个视图中允许并保留第一个视图并阻止手势：

   var gestureView: UIView? = nil

    func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
        if (gestureView == nil || gestureView == touch.view){
            gestureView = touch.view
            return true
        }
        return false
     }

Answer 4

另一种方法是仅比较触摸视图是否为手势视图，因为后代不会通过该条件。一个漂亮，简单的单行：

func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
    return touch.view == gestureRecognizer.view
}

Answer 5

而对于Swift变种：

func gestureRecognizer(gestureRecognizer: UIGestureRecognizer, shouldReceiveTouch touch: UITouch) -> Bool {
    if touch.view.isDescendantOfView(yourSubView){
        return false
    }
    return true
}

很高兴知道，isDescendantOfView返回一个Boolean值，表示接收者是给定视图的子视图还是与该视图相同。

Answer 6

完整的快速解决方案（必须实现委托并为识别器设置）：

class MyViewController: UIViewController UIGestureRecognizerDelegate {

    override func viewDidLoad() {
        let doubleTapRecognizer = UITapGestureRecognizer(target: self, action: #selector(onBaseTapOnly))
        doubleTapRecognizer.numberOfTapsRequired = 2
        doubleTapRecognizer.delegate = self
        self.view.addGestureRecognizer(doubleTapRecognizer)
    }

    func gestureRecognizer(gestureRecognizer: UIGestureRecognizer, shouldReceiveTouch touch: UITouch) -> Bool {
        if touch.view.isDescendantOfView(self.view){
            return false
        }
        return true
    }

    func onBaseTapOnly(sender: UITapGestureRecognizer) {
        if sender.state == .Ended {
            //react to tap
        }
    }
}

Answer 7

使用Swift 5和iOS 12，UIGestureRecognizerDelegate有一个名为gestureRecognizer(_:shouldReceive:)的方法。 gestureRecognizer(_:shouldReceive:)有以下声明：

向代表询问手势识别器是否应该接收表示触摸的对象。

optional func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool

---

下面的完整代码显示了gestureRecognizer(_:shouldReceive:)的可能实现。使用此代码，点击ViewController视图（包括imageView）的子视图将不会触发printHello(_:)方法。

import UIKit

class ViewController: UIViewController, UIGestureRecognizerDelegate {

    override func viewDidLoad() {
        super.viewDidLoad()

        let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(printHello))
        tapGestureRecognizer.delegate = self
        view.addGestureRecognizer(tapGestureRecognizer)

        let imageView = UIImageView(image: UIImage(named: "icon")!)
        imageView.frame = CGRect(x: 50, y: 50, width: 100, height: 100)
        view.addSubview(imageView)

        // ⚠️ Enable user interaction for imageView so that it can participate to touch events.
        // Otherwise, taps on imageView will be forwarded to its superview and managed by it.
        imageView.isUserInteractionEnabled = true
    }

    func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
        // Prevent subviews of a specific view to send touch events to the view's gesture recognizers.
        if let touchedView = touch.view, let gestureView = gestureRecognizer.view, touchedView.isDescendant(of: gestureView), touchedView !== gestureView {
            return false
        }
        return true
    }

    @objc func printHello(_ sender: UITapGestureRecognizer) {
        print("Hello")
    }

}

---

gestureRecognizer(_:shouldReceive:)的另一种实现方式是：

func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
    return gestureRecognizer.view === touch.view
}

但请注意，此替代代码不会检查touch.view是否是gestureRecognizer.view的子视图。

Answer 8

使用CGPoint触摸的变体（SWIFT 4.0）

class MyViewController: UIViewController, UIGestureRecognizerDelegate {

  func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {

// Get the location in CGPoint
    let location = touch.location(in: nil)

// Check if location is inside the view to avoid
    if viewToAvoid.frame.contains(location) {
        return false
    }

    return true
  }
}

Answer 9

清除迅捷的方式

func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
    return touch.view == self.view
}

Answer 10

请注意，gestureRecognizer API已更改为：

gestureRecognizer（_：shouldReceive :)

请特别注意第一个参数外部标签的下划线（跳过）指示器。

使用上面提供的许多示例，我没有收到该事件。下面是一个适用于当前版本的Swift（3+）的示例。

public func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
    var shouldReceive = false
    if let clickedView = touch.view {
        if clickedView == self.view {
            shouldReceive = true;
        }
    }
    return shouldReceive
}

Answer 11

加上上述解决方案，不要忘记检查子视图的User Interaction Enabled。