ES6链接访问先前方法

Question

我对ES6类链接有问题。我有一个如示例中的类。我想访问baz（）方法回调。我想在WorkHard类中使用该回调。我该怎么办？

class WorkHard {

  constructor() {
    this.value_1 = []
    this.value_2 = []
  }
  
  foo(args) {
    this.value_1 = args;
    return this;
  }
  
  baz(cb) {
    this.value_1 = null;
    this.value_2 = null;
    return this;
  }
  
  zoo(args) {
    this.value_2 = args;
    return this;
  }
  
}

const WorkHard = new WorkHard();

WorkHard.foo('TestValue_1').baz(() => {
  console.log('This is baz() method content.');
}).zoo('TestValue_2')

Answer 1

您在第一个参数baz处传递了回调，因此您可以这样调用它：

class WorkHard {
    constructor() {
        this.value_1 = []
        this.value_2 = []
    }

    foo(args) {
        this.value_1 = args;
        return this;
    }

    baz(cb) {
        this.value_1 = null;
        this.value_2 = null;
        
        cb();

        return this;
    }

    zoo(args) {
        this.value_2 = args;
        return this;
    }
}

const workHard = new WorkHard();

workHard
  .foo('TestValue_1')
  .baz(() => {
    console.log('This is baz() method content.');
  })
  .zoo('TestValue_2');

Answer 2

您不会在班级内的任何地方打电话给cb。另外，您的实例不能与该类具有相同的名称。

只需在cb内部执行baz。

class WorkHard {

  constructor() {
    this.value_1 = []
    this.value_2 = []
  }
  
  foo(args) {
    this.value_1 = args;
    console.log('foo called');
    return this;
  }
  
  baz(cb) {
    this.value_1 = null;
    this.value_2 = null;
    cb();
    return this;
  }
  
  zoo(args) {
    console.log('zoo called');
    this.value_2 = args;
    return this;
  }
  
}

const workHard = new WorkHard();

workHard.foo('TestValue_1').baz(() => {
  console.log('This is baz() method content.');
}).zoo('TestValue_2')

如果您希望回调函数控制链接流，请记住，回调函数本身不会返回任何内容，因此您还需要返回该函数调用，而不是返回this

 // inside WorkHard
 baz(cb) {
   this.value_1 = null;
   this.value_2 = null;
   return cb(this);
 }


workHard.foo('TestValue_1').baz((instance) => {
  console.log('Workhard visible variables are', instance);
  return instance;
}).zoo('TestValue_2');