如何在我的 ngOnInit 方法上不使用异步来等待我的呼叫？

Question

I have the following code. To wait for my async call I am using the following

 testi() {
   return new Promise((resolve,reject) => {
     resolve(true);
   })
  }


  async ngOnInit() {
    let pp = await this.testi();
    console.log('pp', pp);

}

is there any other way with which I can wait for the call here without adding async notation on my ngOnInit method?

Answer 1

You can use then instead of await :

testi() {
  return new Promise((resolve,reject) => {
    resolve(true);
  });
}

ngOnInit() {
  this.testi()
    .then(value => console.log(`pp: ${value}`))
}

If you are writing Typescript code, chances are that your compiled code will uses then instead of await . It is the "old" way of using promises.

https://developer.mozilla.org/fr/docs/Web/JavaScript/Reference/Global_Objects/Promise/then

Answer 2

You can have a separated method that handles all the async parts and its result, and within your ngOnInit() you just call that method.

async testi() {
  let pp = await new Promise((resolve,reject) => {
    resolve(true);
  })

  console.log('pp', pp)
}

ngOnInit() {
  this.testi();
}

Answer 3

You can subscribe to it..

 ngOnInit() {
    this.testi().subscribe((pp:any) => {
       console.log('pp', pp);
    }
}