Swift解决方案将于2021年生效。对于这个解决方案，你不需要有对表视图的引用。

func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
    //return !(touch.view is UITableViewCell) <-- doesn't work, the type of the touched class is not UITableViewCell anymore
    
    var v = touch.view
    while v != nil {
        if v is UITableView { return false }
        v = v?.superview
    }
    return true
}

用法:

let gestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(YOUR_METHOD))
gestureRecognizer.delegate = self
YOUR_VIEW.addGestureRecognizer(gestureRecognizer)

我有一个不同的情况，我想要触摸手势函数被调用，只有当用户点击表视图之外。如果用户在表视图内点击，那么触摸手势函数不应该被调用。此外，如果触摸手势函数被调用，它仍然应该将触摸事件传递给被点击的视图，而不是消耗它。

生成的代码是阿卜杜勒拉赫曼·马苏德的答案和尼古拉·尼尔森的答案的组合。

extension MyViewController: UIGestureRecognizerDelegate {
    func addGestureRecognizer() {
        let tapOnScreen = UITapGestureRecognizer(target: self,
                                                action: #selector(functionToCallWhenUserTapsOutsideOfTableView))

        // stop the gesture recognizer from "consuming" the touch event, 
        // so that the touch event can reach other buttons on view.
        tapOnScreen.cancelsTouchesInView = false

        tapOnScreen.delegate = self

        self.view.addGestureRecognizer(tapOnScreen)
    }

    // if your tap event is on the menu, don't run the touch event.
    func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
        if touch.view?.isDescendant(of: self.tableView) == true {
            return false
        }
        return true
    }

    @objc func functionToCallWhenUserTapsOutsideOfTableView() {

        print("user tapped outside table view")
    }
}

在MyViewController类，类有UITableView，在onViewDidLoad()，我确保调用addGestureRecognizer():

class MyViewController: UIViewController {
    ...
    override func viewDidLoad() {
        super.viewDidLoad()
        ...
        self.addGestureRecognizer()
        ...
    }
    ...
}

这是我基于以上答案的解决方案… 这对我很有效……

//Create tap gesture for menu transparent view
UITapGestureRecognizer *rightTableTransparentViewTap = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(rightTableTransparentViewTapMethod:)];
[rightTableTransparentViewTap setCancelsTouchesInView:NO];
[_rightTableTransparentView addGestureRecognizer:rightTableTransparentViewTap];

- (void)rightTableTransparentViewTapMethod:(UITapGestureRecognizer *)recognizer {
    //Write your code here
}

好的，在搜索手势识别器文档后终于找到了。

解决方案是实现UIGestureRecognizerDelegate并添加以下内容:

#pragma mark UIGestureRecognizerDelegate methods
    
- (BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch
{
  if ([touch.view isDescendantOfView:autocompleteTableView]) {
            
    // Don't let selections of auto-complete entries fire the 
    // gesture recognizer
    return NO;
  }
        
  return YES;
}

这样就解决了。希望这也能帮助到其他人。

对于(表/集合)视图下的任何手势，只需将取消视图中的触摸设置为false:

——Interfacebuilder

——代码

<#gestureRecognizer#>.cancelsTouchesInView = false

下面是我的解决方案，它将识别器的shouldReceiveTouch直接绑定到键盘是否显示。

在点击手势识别器委托中:

#pragma mark - UIGestureRecognizerDelegate

- (BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch
{
    if ([PFXKeyboardStateListener sharedInstance].visible) {
        return YES;
    }

    return NO;
}

PFXKeyboardStateListener.h:

@interface PFXKeyboardStateListener : NSObject
{
    BOOL _isVisible;
}

+ (PFXKeyboardStateListener *)sharedInstance;

@property (nonatomic, readonly, getter=isVisible) BOOL visible;

@end

PFXKeyboardStateListener.m:

static PFXKeyboardStateListener *sharedInstance;

@implementation PFXKeyboardStateListener

+ (PFXKeyboardStateListener *)sharedInstance
{
    return sharedInstance;
}

+ (void)load
{
    @autoreleasepool {
        sharedInstance = [[self alloc] init];
    }
}

- (void)dealloc
{
    [[NSNotificationCenter defaultCenter] removeObserver:self];
}

- (BOOL)isVisible
{
    return _isVisible;
}

- (void)didShow
{
    _isVisible = YES;
}

- (void)didHide
{
    _isVisible = NO;
}

- (id)init
{
    if ((self = [super init])) {
        NSNotificationCenter *center = [NSNotificationCenter defaultCenter];
        [center addObserver:self selector:@selector(didShow) name:UIKeyboardDidShowNotification object:nil];
        [center addObserver:self selector:@selector(didHide) name:UIKeyboardWillHideNotification object:nil];
    }
    return self;
}

@end

你可能想要更新键盘监听器的单例模式，我还没有得到它。希望这对每个人都有用，就像对我一样。^ ^