调用函数时不能将网站 link 作为参数包含在内

Question

我使用 request 和 cheerio 库创建了一个脚本，以从网站获取不同的 post 标题及其相应链接。该脚本似乎运行良好。如果您查看下面的脚本，您会发现我使用了 getposts((item,link) => console.log({item,link})); 来调用该函数。

现在，问题是：

How can I include startUrl (website link) as a parameter while calling the function keeping the rest of the logics as they are?

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

const startUrl = 'https://whosebug.com/questions/tagged/web-scraping';

function getposts(callback) {
  request(startUrl, function(error, response, html) {
    if (!error && response.statusCode == 200) {
      var $ = cheerio.load(html);
      $('.summary .question-hyperlink').each(function() {
        var items = $(this).text();
        var links = $(this).attr('href');
        return callback(items, links);
      });
    }
  });
}

getposts((item,link) => console.log({item,link}));

Answer 1

为url创建一个新参数并传入startUrl:

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

const startUrl = 'https://whosebug.com/questions/tagged/web-scraping';

function getposts(url, callback) {
  request(url, function(error, response, html) {
    if (!error && response.statusCode == 200) {
      var $ = cheerio.load(html);
      $('.summary .question-hyperlink').each(function() {
        var items = $(this).text();
        var links = $(this).attr('href');
        return callback(items, links);
      });
    }
  });
}

getposts(startUrl, (item,link) => console.log({item,link}));

调用函数时不能将网站 link 作为参数包含在内

Can't include website link as a parameter while calling a function

javascript

callback

node.js

web-scraping