Question

大家好朋友，我想问问题，我尝试使用标签在视图之间滑动，但我不知道问题？

UISwipeGestureRecognizer *swipe;
swipe = [[UISwipeGestureRecognizer alloc] initWithTarget:self action:@selector(Gest_SwipedLeft:)];
[swipe setDirection:UISwipeGestureRecognizerDirectionLeft];
[label addGestureRecognizer:swipe];
[swipe release];


-(void)Gest_SwipedLeft:(UISwipeGestureRecognizer *) sender{
      ThirdQPage* y=[[ThirdQPage alloc]init];
[self.view.superview addSubview:[y view]];
[self.view removeFromSuperview];}

Answer 1

你可以这样做 -

- (void)hideView:(NSTimer *)timer
{    
    UIView *currentView = (UIView *)[timer userInfo];
    [currentView addSubview:self.yourNewView];
    return;
}

-(void)Gest_SwipedLeft:(UISwipeGestureRecognizer *) sender
{
    [UIView setAnimationTransition:UIViewAnimationTransitionFlipFromLeft 
                           forView:yourCurrentView cache:YES];

    [UIView setAnimationCurve:UIViewAnimationCurveEaseInOut];
    [UIView setAnimationDuration:1];
    [NSTimer scheduledTimerWithTimeInterval:0.5 
                                     target:self 
                                   selector:@selector(hideView:) 
                                   userInfo:yourCurrentView 
                                    repeats:NO];
    [UIView commitAnimations];
}

您可以使用相同的逻辑从newView向oldView向后滑动。

Answer 2

设置label.userInteractionEnabled = YES。 UILabel的默认值为NO，因此会忽略所有触摸。

userInteractionEnabled用于确定是否为用户的布尔值   事件被忽略并从事件队列中删除。

@property（nonatomic，getter = isUserInteractionEnabled）BOOL   userInteractionEnabled

讨论此属性继承自   UIView父类。此类更改此默认值   财产到NO。

使用UISwipeGestureRecognizer滑动视图

2 个答案: