在静态方法(es6)中访问构造函数var [英] access constructor var in static method (es6)

问题描述

遇到以下代码的情况:

class SomeClass{
    constructor(){
        let name="john doe"
    }

    static newName(){
        //i want to get the "name" variable here   
    }
}

在我访问newName()时，在console.log中，无法获取我理解的name变量的引用，调用静态方法时不会实例化该类.所以我想我的问题是，对我来说，调用newName()并访问name变量的最佳方法是什么?我可以在类let name="john doe"上方创建一个变量并以这种方式访问​​它，但我想找出一种将所有内容限制在该类中的方法.

In my console.log when i access newName(), I'm unable to get the reference of name variable which I understand, the class isn't instantiated when I call the static method. So I guess my question is, what would be the best way for me to go about calling newName() and accessing the name variable? I can create a variable above the class let name="john doe" and access it that way, but i'd like to figure out a way to keep everything confined in the class.

推荐答案

首先，让我们暂时忘记static.因此，您的课程应该是这样的:

First off, let's forget about the static for now. So, your class should be like this:

class SomeClass {
  constructor() {
    this.name = "john doe";
  }

  newName() {
    return this.name;
  }
}

看到变量name吗?如果使用let(或var或const)声明它，则它将在constructor中定义为局部变量.因此，它只能在constructor方法内部使用.现在，如果使用关键字this进行设置，它将被定义为实例变量，因此可以在整个类中对其进行访问.

See the variable name? If you declare it with let (or var, or const), it would be defined as local variable in the constructor. Thus, it can only be used inside the constructor method. Now, if you set it with the keyword this, it will be defined as an instance variable, therefore, it can be accessed throughout your class.

现在让我们看看如何实例化您的类并调用方法newName:

Let's see now how you can instantiate your class and call the method newName:

let someClass = new SomeClass(),
    name      = someClass.newName();

如果您真的想使用静态方法，请记住，其中发生的所有事情都不会附加到对象的实例上.

If you really want to use a static method, keep in mind that everything that happens inside it, is not attached to the instance of the object.

您可以在此处阅读更多有关es6类的信息. .

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