javascript 带有请求承诺的异步/等待返回未定义
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Async/Await with Request-Promise returns Undefined
提问by razki
I have two files; server.js and scrape.js, below are the code snippets as they currently stand.
我有两个文件；server.js 和 scrape.js，下面是它们目前的代码片段。
server.js:
服务器.js：
const scrape = require("./scrape");
async function start() {
const response = await scrape.start();
console.log(response);
}
start();
and scrape.js:
和scrape.js：
const cheerio = require("cheerio");
const request = require("request-promise");
go = async () => {
const options = {
uri: "http://www.somewebsite.com/something",
transform: function(body) {
return cheerio.load(body);
}
};
request(options)
.then($ => {
let scrapeTitleArray = [];
$(".some-class-in-html").each(function(i, obj) {
const data = $(this)
.text()
.trim();
scrapeTitleArray.push(data);
});
return scrapeTitleArray;
})
.catch(err => {
console.log(err);
});
};
module.exports = {
start: go
};
So when I spin up server.js, I return undefined to the console.log(response), when I actually want to return the array i've been pushing to, can you see where I'm going wrong?
因此，当我启动 server.js 时，我将 undefined 返回到 console.log(response)，当我实际上想要返回我一直推送到的数组时，你能看出我哪里出错了吗？
回答by Alexander O'Mara
You need to returnsomething from your asyncfunction (a return inside a then does not return from the main function). Either a promise or something you await-ed.
你需要return从你的async函数中得到一些东西（然后在 a 中返回不会从主函数返回）。要么是承诺，要么是你await写过的东西。
Also, make sure to declare your govariable to avoid leaking it into the global space.
另外，请确保声明您的go变量以避免将其泄漏到全局空间中。
const go = async () => {
const options = {
uri: "http://www.somewebsite.com/something",
transform: function(body) {
return cheerio.load(body);
}
};
return request(options)
.then($ => {
let scrapeTitleArray = [];
$(".some-class-in-html").each(function(i, obj) {
const data = $(this)
.text()
.trim();
scrapeTitleArray.push(data);
});
return scrapeTitleArray;
})
.catch(err => {
console.log(err);
});
};
Since you are using an asyncfunction, you might want to take advantage of the awaitsyntax also.
由于您使用的是async函数，因此您可能还想利用await语法。
const go = async () => {
const options = {
uri: "http://www.somewebsite.com/something",
transform: function(body) {
return cheerio.load(body);
}
};
try {
const $ = await request(options);
$(".some-class-in-html").each(function(i, obj) {
const data = $(this)
.text()
.trim();
scrapeTitleArray.push(data);
});
return scrapeTitleArray;
}
catch (err) {
console.log(err);
}
};
回答by Gershom
I believe your gofunction isn't returning any value.
我相信你的go函数没有返回任何值。
You're calling request(options).then(...), but what follows from that promise is never returned by go. I recommend you add a returnstatement:
您正在调用request(options).then(...)，但是从该承诺中得出的内容永远不会被 返回go。我建议你添加一个return声明：
go = async () => {
const options = {
uri: "http://www.somewebsite.com/something",
transform: function(body) {
return cheerio.load(body);
}
};
// The only difference is that it says "return" here:
return request(options)
.then($ => {
let scrapeTitleArray = [];
$(".some-class-in-html").each(function(i, obj) {
const data = $(this)
.text()
.trim();
scrapeTitleArray.push(data);
});
return scrapeTitleArray;
})
.catch(err => {
console.log(err);
});
};
