Question

我有一个缩略图的网格，当其中一个被触摸时，我想显示整个图像。现在，我知道UIImageView不响应触摸事件，但正如this answer中所建议的那样，我创建了一个UIButton来处理事件。请参阅以下代码：

- (void)displayImage
{
        NSUInteger i = 0; //  The actual code is different and works; this is just for brevity's sake.

        // UILazyImageView is a subclass of UIImageView
        UILazyImageView *imageView = [[UILazyImageView alloc] initWithURL:[NSURL URLWithString:[[[self images] objectAtIndex:i] thumbnailUrl]]];

        UIButton *imageButton = [UIButton buttonWithType:UIButtonTypeCustom];
        [imageButton setFrame:[imageView frame]];
        [imageButton addTarget:self action:@selector(imageTapped:) forControlEvents:UIControlEventTouchUpInside];
        [imageView addSubview:imageButton];

        [[self containerView] addSubview:imageView];
        [imageView release];
    }
}

- (void)imageTapped:(id)sender
{
    NSLog(@"Image tapped.");

    // Get the index of sender in the images array 
    NSUInteger index = 0;  // Don't worry, I know.  I'll implement this later

    FullImageViewController *fullImageViewController = [[[FullImageViewController alloc] initWithImageURL:[NSURL URLWithString:[[[self images] objectAtIndex:index] url]]] autorelease];
    [[self navigationController] pushViewController:fullImageViewController animated:YES];
}

确定。因此，我创建了一个自定义按钮，设置其框架以匹配图像框架，告诉它响应内部修饰和如何响应，并将其添加到图像的子视图中。但是当我跑这个时，我什么都没得到。 “图像轻拍”。没有出现在控制台中，所以我知道该消息没有被发送。我在这里做错了什么？

非常感谢你的帮助。

Answer 1

默认情况下，UIImageView的userInteractionEnabled属性设置为NO

添加此行

imageView.userInteractionEnabled = YES;

Answer 2

尝试

imageView.userInteractionEnabled = YES;

这是一个继承自UIView的属性，但根据Apple的文档，UIImageView将其默认值更改为NO，忽略所有事件。

Answer 3

除了设置imageView.userInteractionEnabled = YES之外，您还可以完全取消按钮，并为UITapGestureRecognizer添加UIImageView来处理点按。它看起来像是：

UITapGestureRecognizer *recognizer = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(imageTapped:)];
[imageView addGestureRecognizer:recognizer];

图像视图无法响应触摸事件

3 个答案: