Javascript 如何使用函数的 async-await 从异步函数返回值？

How to return values from async functions using async-await from function?

提问by King Rayhan

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

如何从异步函数返回值？我试着喜欢这个

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,

它返回给我这个，

Promise { <pending> }

采纳答案by The Reason

You cant awaitsomething outside asyncscope. To get expected result you should wrap your console.loginto async IIFE i.e

你不能await做一些超出async范围的事情。要获得预期的结果，您应该将您的内容包装console.log到异步 IIFE 中，即

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

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

Workedsample.

Worked样本。

More information about async/await

有关的更多信息 async/await

Since axiosreturns a promise the async/awaitcan be omitted for the getDatafunction like so:

由于axios返回一个承诺，async/await因此可以getData像这样省略函数：

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

and then do same as we did before

然后像我们以前一样做

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

回答by Pac0

your function getData will return a Promise.

您的函数 getData 将返回一个 Promise。

So you can either:

所以你可以：

• awaitthe function as well to get the result. However, to be able to use await, you need to be in an asyncfunction, 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.)

• await函数也可以得到结果。但是，为了能够使用await，你需要在一个async函数中，所以你需要“包装”这个：

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

  （为了清楚起见，我命名了该函数，但在这种情况下，人们宁愿使用匿名函数调用；请参阅 TheReason 的回答。）

or

或者

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

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

• 将结果用作普通的 Promise，这是异步函数返回的结果。
  您必须then与回调一起使用：

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

回答by Meirion Hughes

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 mainfunction rather than run things in the global scope. i.e.

其他答案已经涵盖了这个罚款；但我想插一句，说养成创建和调用main函数的习惯，而不是在全局范围内运行。IE

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

任何阅读您代码的人都非常清楚这是您的应用程序入口点