Question

在下面的代码中，基于字符串比较，我决定将显示哪个VC。

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现在，所有代码都是重复的，除了类名，我想避免这种情况。但是，swift没有func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) { dataObj = frc.object(at: indexPath) as! Data_Object var pvc: UIViewController? if dataObj.type == "X" { let obj = MainStoryboard().instantiateViewController(withIdentifier: "XVC") as! XVC obj.data = dataObj obj.isFull = true obj.delegate = self pvc = obj as UIViewController } else if dataObj.type == "Y" { let obj = MainStoryboard().instantiateViewController(withIdentifier: "YVC") as! YVC obj.data = dataObj obj.isFull = true obj.delegate = self pvc = obj as UIViewController } else { let obj = MainStoryboard().instantiateViewController(withIdentifier: "ZVC") as! ZVC obj.data = dataObj obj.isFull = true obj.delegate = self pvc = obj as UIViewController } obj.modalPresentationStyle = .popover let popPVC = pvc?.popoverPresentationController popPVC?.sourceView = self.view self.present(pvc!, animated: true, completion: nil) }

我如何做到这一点？我应该使用泛型/模板吗？任何建议！

如果我在这里遗漏了一些傻话，请提前抱歉。

Answer 1

删除重复是显而易见的：

let identifier: String

switch dataObj.type {
    case "X":
       identifier = "XVC"
    case "Y":
       identifier = "YVC"
    default:
       identifier = "ZVC"
}

let pvc = MainStoryboard().instantiateViewController(withIdentifier: identifier) as! PVC
pvc.data = dataObj
pvc.isFull = true
pvc.delegate = self

其中PVC是3个控制器的常见超类，例如：

class PVC : UIViewController {}
class XVC : PVC {}
class YVC : PVC {}
class ZVC : PVC {}

如果您的3个类没有共同的超类，则可以改为使用协议：

protocol PVC : class {
   var data: ...
   var isFul: ...
   weak var delegate: ...
}

并由您的3个班级实施：

简而言之，要删除重复，您需要为3个控制器添加通用接口。使用公共超类或协议。

Answer 2

修改

我更喜欢Sulthan的答案，因为它比我更清洁（不得不承认......）

要完成我的，getControllerFor:type需要返回父类。忘了提它

我编辑了第一个答案来纠正几个未命中并使其更简洁

第一个回答

你可以这样做：

func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) { dataObj = frc.object(at: indexPath) as! Data_Object var pvc: UIViewController? let obj = getControllerFor(type: dateObj.type) obj.data = dataObj obj.isFull = true obj.delegate = self pvc = obj as UIViewController obj.modalPresentationStyle = .popover let popPVC = pvc?.popoverPresentationController popPVC?.sourceView = self.view self.present(pvc!, animated: true, completion: nil) } func getControllerFor(type : String) -> SuperType? { switch type { case "X": return XVC() case "Y": return YVC() case "Z": return ZVC() default: return nil } }

与

class SuperType {} class XVC: SuperType {} class YVC: SuperType {} class ZVC: SuperType {}

理想情况下，type不是String，而是Enum

Answer 3

你可以用枚举很好地完成这项工作。您只需要使每个ViewControllers符合CustomViewController。

enum ViewControllers: String {
    case x = "X"
    case y = "Y"
    case z = "Z"

    func getViewController<T>(dataObj: Data_Object, delegate: YourDelegate?) -> T where T: UIViewController, T: CustomViewController {
        let viewController = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: self.rawValue) as! T
        viewController.data = dataObj
        viewController.isFull = true
        viewController.delegate = delegate

        return viewController
    }
}

protocol CustomViewController: class {
    var data: Data_Object! { get set }
    var isFull: Bool! { get set }
    weak var delegate: YourDelegate? { get set }
}

然后你会像这样使用它：

if dataObj.type == "X" {
     let vc: XVC = ViewControllers.x.getViewController(dataObj: dataObj, delegate: self)
    pvc = vc
}

Answer 4

您可以创建一个获取标识符并返回viewcontroller的函数。由于你的XVC，YVC，ZVC确实有相似之处，你也可以将它们作为其他具有data, isFull,...

属性的VC的子类。

此外，您可以使用与NSClassFromString类似的String(describing: SomeViewController.self)将返回SomeViewController

避免在swift中重复代码

4 个答案: