Ios UIImage?与UIImage不同
我试图将3个图像存储在一个数组中,然后在滚动视图中显示它们,但在这一行中显示了此错误:self.paginatedScrollView?.images=imagesIos UIImage?与UIImage不同,ios,swift,uiscrollview,uiimage,Ios,Swift,Uiscrollview,Uiimage,我试图将3个图像存储在一个数组中,然后在滚动视图中显示它们,但在这一行中显示了此错误:self.paginatedScrollView?.images=images override func viewWillAppear(animated: Bool) { super.viewWillAppear(animated) displayPost(post) paginatedScrollView = PaginatedScrollView(fra
override func viewWillAppear(animated: Bool) {
super.viewWillAppear(animated)
displayPost(post)
paginatedScrollView = PaginatedScrollView(frame: CGRectMake(0, 0, self.view.frame.size.width, self.view.frame.size.height))
self.view.addSubview(paginatedScrollView!) // add to the
let images = [ post!.image1.value, post!.image2.value, post!.image3.value]
self.paginatedScrollView?.images = images
}
这是因为你的形象?是可选的。在使用指定图像之前,只需打开图像的值!接线员 试试这个:
let images: [UIImage] = [ (post!.image1.value)!, (post!.image2.value)!, (post!.image3.value)!]
编辑:您先前创建的数组正在保存选项。所以我只是通过添加一个“!”来打开数组中保存的所有值。阅读此处的选项:这意味着数组正在存储选项。。明确指定数组的类型,并确保在数组中保存未包装的值。哦,这解决了问题!谢谢你能解释一下你做了什么吗!