ios - 使用必要的数据源和委托(delegate)扩展 UIView

Question

我想在 UITableViewCell 中添加自定义的 UIView(将是模拟分段控件)我写了子类`protocol ITISegmentedViewDelegate: class { func segmentedViewButtonChanged(index: Int)

public protocol ITISegmentedViewDataSource : NSObjectProtocol {

@available(iOS 2.0, *)
func segmentedView(itemsInSegmentedView: ITISegmentedView) -> [String]
}

public class ITISegmentedView: UIView {

    var delegate: ITISegmentedViewDelegate?
    var dataSource: ITISegmentedViewDataSource?
    var selectedItem = -1

    override init(frame: CGRect) {
        super.init(frame: frame)

    }

    required public init(coder aDecoder: NSCoder) {
        super.init(coder: aDecoder)!

    self.addButtons()
}

private func addButtons(){
    if delegate == nil || dataSource == nil{
        return
    }

    let height = frame.height
    let width = frame.width

    let array = dataSource!.segmentedView(self)
    let totalItem = array.count

    var startX = CGFloat(0)

    for var index = 0; index < totalItem; ++index{
        let button = UIButton(frame: CGRectMake(startX, 0, width/CGFloat(totalItem), height))
        button.setTitle(array[index], forState: UIControlState.Normal)
        button.tag = index
        button.addTarget(button, action: "onButtonPressed", forControlEvents: .TouchUpInside)
        startX += width/CGFloat(totalItem)
        addSubview(button)
    }

    if totalItem>0{
        selectedItem = 0
        delegate?.segmentedViewButtonChanged(0)
    }
}



func onButtonPressed(button: UIButton){
    if selectedItem != button.tag{
        delegate?.segmentedViewButtonChanged(button.tag)
        selectedItem = button.tag
    }
}

}`

在 Storyboard中添加了 UIView 并设置了类 ITISegmentedView

在我的 ViewController 中:

let cell = tableView.dequeueReusableCellWithIdentifier( cellName, forIndexPath: indexPath)
let seg = (cell.viewWithTag(1) as! ITISegmentedView)
seg.dataSource = self
seg.delegate = self

问题:init(coder aDecoder: NSCoder) 调用 dequeueReusableCell，此时未设置数据源和委托(delegate)，因此 ITISegmentedView 不起作用。

Answer 1

当遇到 nil 时回退到空的 dataSource。

另外，尽量不要使用 !，而是使用 ? 和 ?? 来始终考虑到 Optional 是……好吧。可选。