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

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> }

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 function 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_

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())
})()

[`Working`](

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) sample.

More information about [`async/await`](

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)

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())
})()

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:

```javascript
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][1].)


or

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

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


[1]:

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