我正在使用Promise all同时获取2个URL，但是当我使用await调用这个函数时（因为getAllURLs是异步函数），它给了我一个错误，我该如何解决这个问题呢？
const fetch = require("node-fetch");
let urls = ["https://jsonplaceholder.typicode.com/users","https://jsonplaceholder.typicode.com/users"]
async function getAllUrls(urls) {
try {
var data = await Promise.all(
urls.map((url) => fetch(url).then((response) => response.json()))
);
return data;
} catch (error) {
console.log(error);
throw error;
}
}
const response = await getAllUrls(urls)
console.log(response)
错误:
let responses = await getAllUrls(urls)
await is only valid in async function
您只能在async函数中调用await，例如:
(async () => {
const response = await getAllUrls(urls)
console.log(response)
)()
或者，您可以使用具有顶级await支持的JS引擎或编译器。
