Question

此简化代码：

qtmList = new QTMList();
qtmList.QTMMain = lstQTMmain;
qtmList.QTM = lstQTM;
qtmList.QTMStats = lstQTMStats;

无法解决。

如果将class A { constructor() { this.promise = new Promise(async resolve => { this.resolve = resolve }) } then() { this.promise.then(...arguments) this.resolve(this) } } const x = new A() x.then(res => { console.log(res) })更改为this.resolve(this)，它将起作用。有什么想法吗？

Answer 1

要返回的项目（this）具有.then方法，并且Promise解析器（resolve）看到了这一点，认为您使用Promise对其进行了调用，因此尝试同时解决该承诺：

class A {
  constructor() {
    this.promise = new Promise(async resolve => {
      this.resolve = resolve
      await setTimeout(() => {
        this.x = 1
      }, 1000)
    })
  }

  then() {
    this.promise.then(...arguments)
    this.fin()
    console.log('then running');
  }

  fin() {
    this.resolve(this)
  }
}

const x = new A()
x.then(res => {
  console.log(res)
})

一种可能的解决方案是调用resolve的对象，而该对象包装 this，以使解析器功能看不到.then方法并尝试将其拆开：

class A {
  constructor() {
    this.promise = new Promise(async resolve => {
      this.resolve = resolve
      await setTimeout(() => {
        this.x = 1
      }, 1000)
    })
  }

  then() {
    this.promise.then(...arguments)
    return this.fin();
  }

  fin() {
    this.resolve({ result: this })
  }
}

const x = new A()
x.then(res => {
  console.log(res.result)
})

承诺不能解决这个问题

1 个答案: