在 koa 中发送 HTTP 响应之前如何等待 url 回调?

Question

我有一个koa路由器，我需要调用一个api，其中异步返回结果。这意味着我无法立即获得结果，api 会在正常时调用我的callback url。但是现在我必须像同步 api 一样使用它，这意味着我必须等到调用回调 url。

我的路由器是这样的：

router.post("/voice", async (ctx, next) => {
// call a API here
const params = {
data: "xxx",
callback_url: "http//myhost/ret_callback",
};
const req = new Request("http://xxx/api", {
method: "POST",
body: JSON.stringify(params),
});
const resp = await fetch(req);
const data = await resp.json();
// data here is not the result I want, this api just return a task id, this api will call my url back
const taskid = data.taskid;
// now I want to wait here until I got "ret_callback"
// .... wait .... wait
// "ret_callback" is called now
// get the answer in "ret_callback"
ctx.body = {
result: "ret_callback result here",
}
})

我的回调网址是这样的：

router.post("/ret_callback", async (ctx, next) => {
const params = ctx.request.body;
// taskid will tell me this answer to which question
const taskid = params.taskid;
// this is exactly what I want
const result = params.text;
ctx.body = {
code: 0,
message: "success",
};
})

那么如何才能使这个ayncAPI 像syncAPI 一样呢？

Answer 1

只需将 resolve(( 传递给另一个函数即可。例如，您可以通过以下方式执行此操作：

// use a map to save a lot of resolve()
const taskMap = new Map();
router.post("/voice", async (ctx, next) => {
// call a API here
const params = {
data: "xxx",
callback_url: "http//myhost/ret_callback",
};
const req = new Request("http://xxx/api", {
method: "POST",
body: JSON.stringify(params),
});
const resp = await fetch(req);
const data = await resp.json();
const result = await waitForCallback(data.taskid);
ctx.body = {
result,
} })
const waitForCallback = (taskId) => {
return new Promise((resolve, reject) => {
const task = {};
task.id = taskId;
task.onComplete = (data) => {
resolve(data);
};
task.onError = () => {
reject();
};
taskMap.set(task.id, task);
});
};
router.post("/ret_callback", async (ctx, next) => {
const params = ctx.request.body;
// taskid will tell me this answer to which question
const taskid = params.taskid;
// this is exactly what I want
const result = params.text;
// here you continue the waiting response
taskMap.get(taskid).onComplete(result);
// not forget to clean rubbish
taskMap.delete(taskid);
ctx.body = {
code: 0,
message: "success",
}; })

我没有测试它，但我认为它会起作用。

Answer 2

function getMovieTitles(substr) {
let movies = [];
let fdata = (page, search, totalPage) => {
let mpath = {
host: "jsonmock.hackerrank.com",
path: "/api/movies/search/?Title=" + search + "&page=" + page,
};
let raw = '';
https.get(mpath, (res) => {
res.on("data", (chunk) => {
raw += chunk;
});
res.on("end", () => {
tdata = JSON.parse(raw);
t = tdata;
totalPage(t);
});
});
}
fdata(1, substr, (t) => {
i = 1;
mdata = [];
for (i = 1; i <= parseInt(t.total_pages); i++) {
fdata(i, substr, (t) => {
t.data.forEach((v, index, arrs) => {
movies.push(v.Title);
if (index === arrs.length - 1) {
movies.sort();
if (parseInt(t.page) === parseInt(t.total_pages)) {
movies.forEach(v => {
console.log(v)
})
}
}
});
});
}
});

}
getMovieTitles("tom")

Answer 3

好的，首先，这对你来说不应该是一个"目标"。NodeJS作为ASync工作得更好。

但是，让我们假设您出于某种原因仍然想要它，因此请查看 npm 上的同步请求包(那里有一个巨大的注释，您不应该在生产中这样做。

但是，我希望你的意思是如何使这个 API 更简单(就像在一个调用中有点事情一样(。您仍然需要.next或await但无论如何都会是一个电话。
如果是这种情况，请对此答案发表评论，我可以给你写一个我自己使用的方法。

Answer 4

这个怎么样？

router.post("/voice", async (ctx, next) => {
const params = {
data: "xxx",
callback_url: "http//myhost/ret_callback",
};
const req = new Request("http://xxx/api", {
method: "POST",
body: JSON.stringify(params),
});
const resp = await fetch(req);
const data = await resp.json();
// data here is not the result I want, this api just return a task id, this api will call my url back
const taskid = data.taskid;
let response = null;
try{
response = await new Promise((resolve,reject)=>{
//call your ret_callback and when it finish call resolve(with response) and if it fails, just reject(with error);
});
}catch(err){
//errors
}
// get the answer in "ret_callback"
ctx.body = {
result: "ret_callback result here",
}
});