Question

我遇到错误

“ UIScrollView”类型的值？没有成员“ delagate”

我一直在youtube上关注本教程 https://www.youtube.com/watch?v=AgUubgI-ZjI&feature=share

我该如何解决？

@IBOutlet weak var pageControl:UIPageControl!
@IBOutlet weak var scrollView:UIScrollView!

var images: [String] = ["About","Fidget Balancer","Home"]
var frame = CGRect(x:0,y:0,width:0,height:0)

override func viewDidLoad() {
    super.viewDidLoad()
    // Do any additional setup after loading the view, typically from a nib.
    pageControl.numberOfPages = images.count
    for index in 0..<images.count {
        frame.origin.x = scrollView.frame.size.width * CGFloat(index)
        frame.size = scrollView.frame.size

        let imgView = UIImageView(frame: frame)
        imgView.image = UIImage(named: images[index])
        self.scrollView.addSubview(imgView)
    }
    scrollView.contentSize = CGSize(width: (scrollView.frame.size.width * CGFloat(images.count)), height: scrollView.frame.size.height)
    scrollView.delagate = self
}

//Scrollview Method
// ==============================
func scrollViewDidEndDecelerating(_ scrollView: UIScrollView) {
    var pageNumber = scrollView.contentOffset.x / scrollView.frame.size.width
    pageControl.currentPage = Int (pageNumber)
}

Answer 1

替换

scrollView.delagate = self

使用

scrollView.delegate = self

并确保您的班级符合UIScrollViewDelegate

Answer 2

没有词偏词，它称为委托 https://developer.apple.com/documentation/uikit/uiscrollviewdelegate

类型为“ UIScrollView”的值？没有成员“ delagate”

2 个答案: