3

使用 UIViews 和 UIImageViews 数组([[[UIApplication sharedApplication] window] subviews])。我只需要删除 UIImageView 类型的最高索引的对象。

4

3 回答 3

6

您可以使用indexOfObjectWithOptions:passingTest:方法反向搜索数组以查找使用块通过测试的对象,然后删除结果位置的对象:

NSUInteger pos = [myArray indexOfObjectWithOptions:NSEnumerationReverse
                          passingTest:^(id obj, NSUInteger idx, BOOL *stop) {
    return [obj isKindOfClass:[UIImageView class]]; // <<== EDIT (Thanks, Nick Lockwood!)
}];
if (pos != NSNotFound) {
    [myArray removeObjectAtIndex:pos];
}
于 2012-02-03T15:36:27.120 回答
5

另一个基于块的解决方案

[window.subviews enumerateObjectsWithOptions:NSEnumerationReverse 
                                  usingBlock:^(id view, NSUInteger idx, BOOL *stop) 
    {
        if ([view isKindOfClass:[UIImageView class]]){
            [view removeFromSuperview];
            *stop=YES;
    }
}];

非阻塞解决方案:

for (UIView *view in [window.subview reverseObjectEnumerator])
{
    if ([view isKindOfClass:[UIImageView class]]){
            [view removeFromSuperview];
            break;
    }
}

我发布了一些演示代码,显示了两种解决方案。

于 2012-02-03T15:45:57.233 回答
3

怎么样:

UIWindow *window = [[UIApplication sharedApplication] window];
UIView *imageView = nil;
for (UIView *view in window.subviews)
{
    if ([view isKindOfClass:[UIImageView class]])
    {
        imageView = view;
    }
}

//this will be the last imageView we found
[imageView removeFromSuperview];
于 2012-02-03T15:34:52.067 回答