如何在Node.js中等待回调函数调用?

Question

我是Node.js和Javascript的新手，我使用npm包重试发送请求到服务器。

const retry = require('retry');
async function HandleReq() {
//Some code
return await SendReqToServer();
}
async function SendReqToServer() {

operation.attempt(async (currentAttempt) =>{
try {
let resp = await axios.post("http://localhost:5000/api/", data, options);
return resp.data;
} catch (e) {
if(operation.retry(e)) {throw e;}
}
});
}

我得到空响应，因为在传递给operation.attempt的函数解析承诺之前，SendReqToServer返回一个承诺。

如何解决这个问题?

Answer 1

这个问题的解取决于operation.attempt。如果它返回一个承诺，您也可以简单地在SendReqToServer中返回该承诺。但是通常带回调的异步函数不会返回承诺。创建你自己的承诺:

const retry = require('retry');
async function HandleReq() {
//Some code
return await SendReqToServer();
}
async function SendReqToServer() {

return new Promise((resolve, reject) => {
operation.attempt(async (currentAttempt) => {
try {
let resp = await axios.post("http://localhost:5000/api/", data, options);
resolve(resp.data);
return resp.data;
} catch (e) {
if(operation.retry(e)) {throw e;}
}
});
});
}

Answer 2

如果函数中没有错误，返回operation.attempt()将返回resp.data到sendReqToServer()。目前，您只是将resp.data返回到operation.attempt()。您还需要返回operation.attempt()

const retry = require('retry');
async function HandleReq() {
//Some code
return SendReqToServer();
}
async function SendReqToServer() {

return operation.attempt(async (currentAttempt) => {
try {
let resp = await axios.post("http://localhost:5000/api/", data, options);
return resp.data;
} catch (e) {
if(operation.retry(e)) {throw e;}
}
});
}