Question

我试图弄清楚如何在请求中按顺序运行回调函数。目前，我的循环运行了10次但是在请求完成之前不等待回调函数完成，然后才转到下一次迭代。我的输出与我所期望的完全不同，我不确定为什么某些东西会被打印出来。以下是我的代码截至目前的情况：

var express = require('express');
var path = require('path');
var request = require('request');
var cheerio = require('cheerio');
var fs = require('fs');
var app = express();
var url;
for(var i=0; i < 10; i++ ){
    url = "http://www.ratemyprofessors.com/ShowRatings.jsp?tid=" + i;
    request(url, function(err, resp, body){
        console.log("hello");
        var $ = cheerio.load(body);
        if($('.error').text().substring(0, 14) == "Page Not Found"){
            console.log("sorry page not found");
            return;
        }else{
        console.log($('.error').text().substring(0, 14) );
        var pfname = $('.pfname');
        var plname = $('.plname');
        var professorName = pfname.text().replace(/\s/g, '') + " " +plname.text().replace(/\s/g, '');
        console.log(professorName);
        console.log(url);
        return;
        }

    });
}

以下是我得到的输出：

    hello
    sorry page not found
    hello
    sorry page not found
    hello
    sorry page not found
    hello
    sorry page not found
    hello
    sorry page not found
    hello
    sorry page not found
    hello
    sorry page not found
    hello

    Michael Beeson
    http://www.ratemyprofessors.com/ShowRatings.jsp?tid=9
    hello

    Sami Khuri
    http://www.ratemyprofessors.com/ShowRatings.jsp?tid=9
    hello

    aaa aaa
    http://www.ratemyprofessors.com/ShowRatings.jsp?tid=9

Here is the proper output:


aaa aaa
http://www.ratemyprofessors.com/ShowRatings.jsp?tid=1

Sami Khuri
http://www.ratemyprofessors.com/ShowRatings.jsp?tid=2


Michael Beeson
http://www.ratemyprofessors.com/ShowRatings.jsp?tid=3

sorry page not found
sorry page not found
sorry page not found
sorry page not found
sorry page not found
sorry page not found
sorry page not found

Answer 1

您的代码中存在多个问题，但主要问题是您在for循环内运行异步操作，因此您的for循环将启动所有异步操作，然后他们将逐一完成。循环调用之间共享的任何变量都会相互争吵。

所以，简而言之，我做了：

删除了所有共享变量，因此每个循环调用都有自己的变量（没有冲突）。
切换到请求承诺，以便我们可以使用Promise.all()更轻松地告诉我们什么时候完成。
从每个.then()处理程序返回我们想要的值，以便Promise.all()收集该值作为每次循环调用的最终值。
因为似乎没有理由对您的操作进行排序，我让它们全部并行运行（速度更快），然后让Promise.all()将结果放入最终数组中结果。

以下是代码：

const express = require('express');
const path = require('path');
const rp = require('request-promise');
const cheerio = require('cheerio');
const fs = require('fs');
const app = express();

let promises = [];
for (let i = 0; i < 10; i++ ) {
    let url = "http://www.ratemyprofessors.com/ShowRatings.jsp?tid=" + i;
    promises.push(rp(url).then(function(body) {
        console.log(url);
        let $ = cheerio.load(body);
        if($('.error').text().substring(0, 14) == "Page Not Found"){
            console.log("sorry page not found");
            return null;
        } else {
            console.log($('.error').text().substring(0, 14) );
            let pfname = $('.pfname');
            let plname = $('.plname');
            let professorName = pfname.text().replace(/\s/g, '') + " " +plname.text().replace(/\s/g, '');
            console.log(professorName);
            return professorName;
        }

    }));
}

// see when they are all done
Promise.all(promises).then(results => {
    // array of results, some entries that were not found may be null
    console.log(results);
}).catch(err => {
    console.log(err);
});

如果你想一次一个地对它们进行排序，那么第二个请求在第一个请求完成之前就不会开始，可以使用async/await这样做：

const express = require('express');
const path = require('path');
const rp = require('request-promise');
const cheerio = require('cheerio');
const fs = require('fs');
const app = express();

async function run() {
    let results = [];
    for (let i = 0; i < 10; i++ ) {
        let url = "http://www.ratemyprofessors.com/ShowRatings.jsp?tid=" + i;
        try {
            let body = await rp(url);
            console.log("hello");
            let $ = cheerio.load(body);
            if($('.error').text().substring(0, 14) == "Page Not Found"){
                console.log("sorry page not found");
                results.push(null);
            } else {
                console.log($('.error').text().substring(0, 14) );
                let pfname = $('.pfname');
                let plname = $('.plname');
                let professorName = pfname.text().replace(/\s/g, '') + " " +plname.text().replace(/\s/g, '');
                console.log(professorName);
                console.log(url);
                results.push(professorName);
            }

        } catch(e) {
            console.log(url, e);
            results.push(null);
        }
    }
    return results;
}

run().then(results => {
    console.log(results);
}).catch(err => {
    console.log(err);
});

nodejs请求最后运行

1 个答案: