Question

我想抓取youtube频道，我列出了1500个频道列表。我想通过节点一一刮掉它。问题出在循环中，我想停止循环，直到获得当前通道的响应和处理响应为止。

异步并等待

const cheerio = require('cheerio');
const request = require('request');
var fs = require('fs');

async function getTitle(){
    var data =  await fs.readFileSync('./channels.json');
    var Parsed_Data = await JSON.parse(data);
    console.log(Parsed_Data);
    for (x of Parsed_Data){
        // var url = 'https://www.youtube.com/channel/' + x.id + '/about';
        // console.log(url);
        request( 'https://www.youtube.com/channel/' + x.id + '/about', function(err, res, body) {
        let $ =  cheerio.load(body);
        console.log($);
        let title =  $('title');
        console.log(title.text());
        }

    )};

想要每个频道的标题

Answer 1

您可以这样做：

function getBody(url) {
    return new Promise((resolve, reject) => {
        request( 'https://www.youtube.com/channel/' + x.id + '/about', function(err, res, body) {
            if (err) {
                return reject(err);
            }
            resolve(body);
        });
    })
}

async function getTitle(){
    var data =  await fs.readFileSync('./channels.json');
    var Parsed_Data = await JSON.parse(data);
    console.log(Parsed_Data);
    for (x of Parsed_Data){
        const body = await getBody('https://www.youtube.com/channel/' + x.id + '/about');

        let $ =  cheerio.load(body);
        console.log($);
        let title =  $('title');
        console.log(title.text());

)};

Answer 2

首先，安装请求承诺：

npm i request-promise

然后，将代码更改为此：

const cheerio = require('cheerio');
const rp = require('request-promise');
var fs = require('fs');

async function getTitle(){
    var data =  await fs.readFileSync('./channels.json');
    var Parsed_Data = JSON.parse(data);
    console.log(Parsed_Data);
    for (x of Parsed_Data){
        // var url = 'https://www.youtube.com/channel/' + x.id + '/about';
        // console.log(url);
        let body=await rp( 'https://www.youtube.com/channel/' + x.id + '/about', 
        let $ =  cheerio.load(body);
        console.log($);
        let title =  $('title');
        console.log(title.text());
    }
}

实际上，您不需要在JSON.parse（data）;旁边等待。

保持循环，直到我从API引起共鸣

2 个答案: