Question

我想获取父类名（import Foundation import UIKit import Darwin class View3on3 : UIViewController, UITextFieldDelegate { @IBOutlet weak var APTeams: UITextField! @IBOutlet weak var APRounds: UITextField! @IBOutlet weak var APBreakers: UITextField! @IBOutlet weak var ToAPBreak: UIButton! override func viewDidLoad() { super.viewDidLoad() initializeTextFields() APTeams.delegate = self APRounds.delegate = self APBreakers.delegate = self ToAPBreak.backgroundColor = UIColor.clearColor() ToAPBreak.layer.cornerRadius = 11 ToAPBreak.layer.borderWidth = 1 ToAPBreak.layer.borderColor = self.view.tintColor.CGColor } func initializeTextFields() { APTeams.keyboardType = UIKeyboardType.NumberPad APRounds.keyboardType = UIKeyboardType.NumberPad APBreakers.keyboardType = UIKeyboardType.NumberPad } func textFieldShouldReturn(textField: UITextField) -> Bool { if (textField == APTeams){ APRounds.becomeFirstResponder() } else if (textField == APRounds){ APBreakers.becomeFirstResponder() } else if (textField == APBreakers){ APBreakers.resignFirstResponder() } else { //etc// } return true }），但我只能使用此代码（Parent）检索子类名...

＆＃13;

Child

＆＃13;

有可能吗？

谢谢！

Answer 1

ES6类相互继承。因此，当Child引用Object.getPrototypeOf(instance.constructor)时，您可以使用Parent获取Object.getPrototypeOf(instance.constructor).name == "Parent";，然后访问.name：

{{1}}

当然，完整的ES6合规性和非缩小代码是必不可少的。你不应该依赖代码中的函数名。

Answer 2

这是有趣的事情：

class J {}
    
class K extends J {}

class L extends K {}

function getBaseClass(targetClass){
  if(targetClass instanceof Function){
    let baseClass = targetClass;

    while (baseClass){
      const newBaseClass = Object.getPrototypeOf(baseClass);

      if(newBaseClass && newBaseClass !== Object && newBaseClass.name){
        baseClass = newBaseClass;
      }else{
        break;
      }
    }

    return baseClass;
  }
}

console.log(getBaseClass(L)); // Will return J.

Answer 3

你可以在技术上做到

// instanceProto === Child.prototype
var instanceProto = Object.getPrototypeOf(instance);

// parentProto === Parent.prototype
var parentProto = Object.getPrototypeOf(instanceProto);
console.log(parentProto.constructor.name);

请记住，这些名称可能都会被缩小器破坏。

Answer 4

来自proto属性：

Child['__proto__'].name

此处的示例：https://stackblitz.com/edit/typescript-s5brk9

Answer 5

另一种简单的解决方案：

class Foo {}

class Bar extends Foo{}

Object.getPrototypeOf(Bar) === Foo // true

Bar.__proto__ === Foo // true

console.log(Bar.__proto__.name) // Foo

从ES6的孩子那里获取父类名？

5 个答案: