静态方法中的访问构造函数var（es6）

Question

Running into a situation where I have the following code:

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

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

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.

Answer 1

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;
  }
}

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.

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

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

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.

You can read more about es6 classes here .