如何将子视图设置为第一响应者？

Question

我试图在单击关闭按钮时关闭图像视图。但是按钮不响应触摸事件。 这是我的代码：

-(void)imageTouchUP:(id)sender
{
    UIImageView *imageView=[[UIImageView alloc]initWithFrame:CGRectInset(self.view.bounds, 30 , 30)];
    imageView.image=self.testimonialImage;
    imageView.tag=1;
    UIButton *closeButton=[UIButton buttonWithType:UIButtonTypeCustom];
    closeButton.frame=CGRectMake(0, 0, 20, 20);
    [closeButton setImage:[UIImage imageNamed:@"close.jpg"] forState:UIControlStateNormal];
    [closeButton addTarget:self action:@selector(buttonClicked:) forControlEvents:UIControlEventTouchUpInside];
    [imageView addSubview:closeButton];
    [self.view addSubview:imageView];
}

-(void)buttonClicked:(id)sender
{
    UIView *subView=[self.view viewWithTag:1];
    [subView removeFromSuperview];
}

Answer 1

为了使UIImageView处理触摸，您需要将userInteractionEnabled设置为YES，它默认为NO。 尝试：

[imageView setUserInteractionEnabled:YES]; 

参考

Answer 2

UIImageView userInteractionEnabled有1个属性，其值默认为NO。 所以在你的代码中添加

UIImageView *imageView=[[UIImageView alloc]initWithFrame:CGRectInset(self.view.bounds, 30 , 30)];
imageView.userInteractionEnabled = YES;

Answer 3

也许在您的self.view中还有另一个带有tag = 1的视图。 请尝试编辑这样单击的按钮：

- (void) buttonClicked:(UIButton *)btn
{
    UIView *view = btn.superview;
    if ([view isKindOfClass:[UIImageView class]]) {
        [view removeFromSuperview];
    }
}