CODEWITHSUNDEEP
javascriptnode.jsweb-scrapingcheerio
Aashish Kumar
Aashish Kumar

Reputation: 115

Web Scraping with cheerio Nodejs

I'm trying to scrape the jobs with zero experience on MonsterIndia.com, so i wrote the following code using cheerio and nodejs, i observed that i can search for php jobs by seaching it like https://www.monsterindia.com/**php**-jobs.html but if i want to search the php jobs with zero experience i had to add filters manually on the site, but it does not reflect in the url of the page, so how can i achieve this, i am a complete beginner in web scraping, please help.

var request = require('request');
var cheerio = require('cheerio');
const context = "php";
function scraper(context){
    request('http://www.monsterindia.com/'+context+"-jobs.html", function (error, response, html) {
        if (!error && response.statusCode == 200) {
            console.log("Request Called");
            var $ = cheerio.load(html);
            var jobs = [];
            var json = {title : "", link:"", description:"", };
            $('a.title_in').each(function(i , element){


                console.log($(this).attr('title'));

            })
        } 
        if(error){
            console.log(error);
        } 

    });
}
scraper(context);

Upvotes: 1

Views: 961

Answers (2)

Aashish Kumar
Aashish Kumar

Reputation: 115

The best approach is to use .filter() function in cheerio, following is my code which implements one such filters.

var job = $('div.job-container').filter(function (i, el) {
 var exp = $(this).children('div.view-apply- 
  container').children('div.padding-top-5').children('div.col-md- 
  3.col-xs-3.col-lg-3').children('span.experience');
  if (!exp){
      return $('div.job-container');
  }
  else{
      exp = exp.text().charAt(0) === '0';
      return exp;
  }
});

Upvotes: 0

Pezhman Parsaee
Pezhman Parsaee

Reputation: 1249

You can use casperjs library to achieving your aim. It is a simple example that are placed in its website to searching for a word using Google search engine and then fetching links that is placed in the first page of search result.

var links = [];
var casper = require('casper').create();

function getLinks() {
    var links = document.querySelectorAll('h3.r a');
    return Array.prototype.map.call(links, function(e) {
        return e.getAttribute('href');
    });
}

casper.start('http://google.fr/', function() {
   // Wait for the page to be loaded
   this.waitForSelector('form[action="/search"]');
});

casper.then(function() {
   // search for 'casperjs' from google form
   this.fill('form[action="/search"]', { q: 'casperjs' }, true);
});

casper.then(function() {
    // aggregate results for the 'casperjs' search
    links = this.evaluate(getLinks);
    // now search for 'phantomjs' by filling the form again
    this.fill('form[action="/search"]', { q: 'phantomjs' }, true);
});

casper.then(function() {
    // aggregate results for the 'phantomjs' search
    links = links.concat(this.evaluate(getLinks));
});

casper.run(function() {
    // echo results in some pretty fashion
    this.echo(links.length + ' links found:');
    this.echo(' - ' + links.join('\n - ')).exit();
});

Upvotes: 0

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