ios - UILongPressGestureRecognizer iOS 用于连续 Action

Question

我有一个按钮，我想“按下并按住”按钮，这样它就会一直打印“长按”，直到我松开按键为止。

我在 ViewDidLoad 中有这个:

[self.btn addTarget:self action:@selector(longPress:) forControlEvents:UIControlEventTouchDown];

和

- (void)longPress: (UILongPressGestureRecognizer *)sender {
    if (sender.state == UIControlEventTouchDown) {
      NSLog(@"Long Press!");
    }
}

我也试过这个:

 UILongPressGestureRecognizer *lpgr = [[UILongPressGestureRecognizer alloc]initWithTarget:self action:@selector(longPress:)];
 lpgr.minimumPressDuration = 0.1;
 lpgr.numberOfTouchesRequired = 1;
 [self.btn addGestureRecognizer:lpgr];

它只会打印长按!即使我按住按钮也有一次。谁能告诉我哪里做错了或错过了什么？谢谢!

Answer 1

首先，UIButton 的目标-操作对回调仅在相应事件触发时执行一次。

UILongPressGestureRecognizer 需要 minimumPressDuration 才能进入 UIGestureRecognizerStateBegan 状态，然后当手指移动时，回调将触发 UIGestureRecognizerStateChanged 状态。最后，释放手指时的UIGestureRecognizerStateEnded状态。

您的要求是按下按钮时反复触发。一流的都不能满足你的需要。相反，您需要在按下按钮时设置一个重复触发计时器，并在按钮释放时释放它。

所以代码应该是:

    [btn addTarget:self action:@selector(touchBegin:) forControlEvents: UIControlEventTouchDown];
    [btn addTarget:self action:@selector(touchEnd:) forControlEvents:
        UIControlEventTouchUpInside | UIControlEventTouchUpOutside | UIControlEventTouchCancel];

- (void)touchBegin:(UIButton*)sender {
    _timer = [NSTimer scheduledTimerWithTimeInterval:0.1 target:self selector:@selector(longPress:) userInfo:nil repeats:YES];
}

- (void)touchEnd:(UIButton*)sender {
    [_timer invalidate];
    _timer = nil;
}

- (void)longPress:(NSTimer*)timer {
    NSLog(@"Long Press!");
}

顺便说一句，UIButton 和 UILongPressGestureRecognizer 都没有类型为 UIControlEvents

的状态