異步 function 在返回之前不等待等待

Question

考慮以下代碼：

 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. 有沒有解決的辦法？

Answer 1

您需要await f() 。 這是一個例子：

 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

您需要等待 f function 返回的 promise。 這就是為什么它打印 promise 而不是結果的原因。

console.log(await f());