异步 function 在返回之前不等待等待

Question

Consider the following code:

 const getName = () => new Promise(resolve => setTimeout(resolve, 1000, 'xxx')); f = async () => { let name = await getName(); console.log(name); return name; } console.log(f());

The function will wait before printing "name" but it will still return the promise instead of the result and won't print the correct output outside the function. Is there a way around this?

Answer 1

You need to await for f() . Here is an example:

 const getName = () => new Promise(resolve => setTimeout(resolve, 1000, 'xxx')); const f = async () => { const name = await getName(); console.log("1. "+name); return name; } const run = (async () => { const ret = await f(); console.log("2. "+ret); })();

Answer 2

You need to await the promise returned by the f function. That is why its printing a promise instead of a result.

console.log(await f());