Question

有没有办法隐藏UIButton直到按下UIImageView？当图片被按下时，我需要显示后退按钮，就像它在iPhone上的照片应用程序中工作一样？这是我的UIButton的代码：

- (void)viewDidLoad {
    [super viewDidLoad];

    [self ladeImage];
    UIButton *btn = [UIButton buttonWithType:UIButtonTypeRoundedRect];
    btn.frame = CGRectMake(10, 10, 40, 40);
    [btn addTarget:self action:@selector(goToViewA) forControlEvents:UIControlEventTouchUpInside];
    [btn setTitle:@"<<" forState:UIControlStateNormal];
    [self.view addSubview:btn];

    }

Answer 1

第一步：btn.hidden = YES

然后你必须继承UIImageView以对其touchesEnded:事件作出反应，并在那里更改按钮的隐藏属性。为此，正确的方法是创建一个协议（使用viewTouched方法）。在包含您的按钮和ImageView的viewController中实现该协议。将一个委托属性添加到子类ImageView（即id<MyCustomProtocol> _delagate;）并将视图控制器分配给此属性。

Answer 2

btn.hidden = YES;

UIImageView *imageView = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"image name"]];
imageView.userInteractionEnabled = YES; // here to enable touch event
UITapGestureRecognizer *tap = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(handleTapGestureRecongizer:)]; // handleTapGestureRecongizer is method will call when tap even fire
[imageView addGestureRecognizer:tap]; // Add Tap gesture recognizer to image view
[tap release], tap = nil;
[self.view addSubview:imageView];
[imageView release], imageView = nil;

方法handlerTapGestureRecognizer：

- (void)handleTapGestureRecongizer:(UITapGestureRecognizer *)gestureRecognizer{
if (gestureRecognizer.state == UIGestureRecognizerStateEnded) {
    btn.hidden = NO;
}

}

玩得开心！

触摸图像时显示UIButton？

2 个答案: