ios - 如何在突出显示(点击)时使 UIImageView 变暗？

Question

我有一个显示图片的 UIImageView。点击并按住时，只要用户将手指放在 ImageView 上，我就希望它变暗。基本上，我希望它像 UIButton 一样工作。

我目前使用 UIGestureRecognizer 来执行此操作:

- (void)viewDidLoad
{
    [super viewDidLoad];

    UITapGestureRecognizer *singleTap = [[UITapGestureRecognizer alloc]
                                         initWithTarget:self
                                         action:@selector(pictureViewTapped)];
    singleTap.numberOfTapsRequired = 1;
    singleTap.numberOfTouchesRequired = 1;
    [self.pictureImageView addGestureRecognizer:singleTap];
    [self.pictureImageView setUserInteractionEnabled:YES];
}

在我的 pictureViewTapped 中，我有这个(现在):

- (void)pictureViewTapped {
    NSLog(@"Picture view was tapped!");
}

如何使 UIImageView 变暗？感谢您的帮助。

Answer 1

最简单的方法是在 QuartzCore 框架中使用层:

在您的项目设置中添加:QuartzCore.framework 到项目中添加到 .h 文件:#import

为长按创建手势，而不是像您描述的那样点击该行为:

 self.longTap = [[UILongPressGestureRecognizer alloc] initWithTarget:self action:@selector(darkenImage)];
    [self.imageView setUserInteractionEnabled:YES];
    [self.imageView addGestureRecognizer:self.longTap];

手势激活时调用的方法:

-(void)darkenImage {
    switch (self.longTap.state) {
        case UIGestureRecognizerStateBegan: // object pressed
        case UIGestureRecognizerStateChanged:
            [self.imageView.layer setBackgroundColor:[UIColor blackColor].CGColor];
            [self.imageView.layer setOpacity:0.9];
            break;
        case UIGestureRecognizerStateEnded: // object released
            [self.imageView.layer setOpacity:1.0];
            break;
        default: // unknown tap
            NSLog(@"%i", self.longTap.state);
            break;
    }
}