我如何从一个具有某些类似变量的视图控制器列表中声明一个视图控制器,而无需重复自己和而不创建基本的视图控制器?像这样:
let storyboard = UIStoryboard(name: "MainStoryboard", bundle: nil)
swtich(type){
case "A":
vcGeneric = storyboard.instantiateViewController(withIdentifier: "TypeAViewController") as! TypeAViewController
break;
case "B":
vcGeneric = storyboard.instantiateViewController(withIdentifier: "TypeBViewController") as! TypeBViewController
break;
}
vcGeneric.variableSame1 = "SomeValue1"
vcGeneric.variableSame2 = "SomeValue2"
vcGeneric.variableSame3 = "SomeValue3"
self.present(vcGeneric, animated: true, completion: nil)
我通过声明var vcGeneric: UIViewController
进行了尝试,但收到编译错误Value of type "UIViewController" has no member "variableSame1"
答案 0 :(得分:3)
使用协议定义通用属性,如下所示:
protocol CommonViewController {
var variableSame1: String? { get set }
var variableSame2: String? { get set }
var variableSame3: String? { get set }
}
class ViewController1: UIViewController, CommonViewController {
var variableSame1: String?
var variableSame2: String?
var variableSame3: String?
}
class ViewController2: UIViewController, CommonViewController {
var variableSame1: String?
var variableSame2: String?
var variableSame3: String?
}
class ViewController3: UIViewController, CommonViewController {
var variableSame1: String?
var variableSame2: String?
var variableSame3: String?
}
let storyboard = UIStoryboard(name: "MainStoryboard", bundle: nil)
var type = "A"
var vcGeneric: CommonViewController?
switch type {
case "A":
vcGeneric = storyboard.instantiateViewController(withIdentifier: "TypeAViewController") as! ViewController1
break;
case "B":
vcGeneric = storyboard.instantiateViewController(withIdentifier: "TypeBViewController") as! ViewController2
break;
default:
vcGeneric = storyboard.instantiateViewController(withIdentifier: "TypeCViewController") as! ViewController3
break;
}
vcGeneric.variableSame1 = "SomeValue1"
vcGeneric.variableSame2 = "SomeValue2"
vcGeneric.variableSame3 = "SomeValue3"
self.present(vcGeneric, animated: true, completion: nil)