How to return values from async functions using as

问题:

How can I return the value from an async function? I tried to like this

const axios = require('axios');
async function getData() {
    const data = await axios.get('https://jsonplaceholder.typicode.com/posts');
    return data;
}
console.log(getData());

it returns me this,

Promise { <pending> }

回答1:

You cant await something outside async scope. To get expected result you should wrap your console.log into async IIFE i.e

async function getData() {
  return await axios.get('https://jsonplaceholder.typicode.com/posts');
}

(async () => {
  console.log(await getData())
})()

Worked sample.

More information about async/await

Since axios returns a promise the async/await can be omitted for the getData function like so:

function getData() {
  return axios.get('https://jsonplaceholder.typicode.com/posts');
}

and then do same as we did before

(async () => {
   console.log(await getData())
})()

回答2:

The other answers have covered this fine; but I'd like to chip in and say get in the habit of creating and calling a main function rather than run things in the global scope. i.e.

async main(){
  let result = await getData();
}

main().catch(console.log);

This is pretty clear to anyone reading your code that this is your app entry point

回答3:

your function getData will return a Promise.

So you can either:

• await the function as well to get the result. However, to be able to use await, you need to be in an async function, so you need to 'wrap' this:

  async function callAsync() {
     var x = await getData();
     console.log(x);
  }
  callAsync();
  

  (I named the function for sake of clarity, but in this scenario, one would rather use an anonymous function call; see TheReason's answer.)

or

• use the result as a normal Promise, which is what an async function returns.
  You have to use then with a callback:

  getData().then(x => { 
      console.log(x); 
  }