How can I open multiple links on a webpage and scrape there data?

Question

I hope you guys are best with your health and R&D work.

import webbrowser
import scrapy
from urllib.request import urlopen
import re
from scrapy.selector import Selector

class QuotesSpider(scrapy.Spider):
    name = "forum"

    def start_requests(self):
        urls = ['https://tribune.com.pk/'], #'https://www.siasat.pk/forum/content.php/', 'http://hamariweb.com/news/', 'https://www.urdupoint.com/pakistan/all-news/']
        for url in urls:
            website = urlopen(url)
            webbrowser.open(website)
            print("HELLO WORLD")
            html = website.read()
            all_links = re.findall('"((http|ftp)s?://.*?)"', html)
            for link in all_links:
                yield scrapy.Request(url=link, callback=self.parse)
            
                        
    def parse(self, response):
        page = response.url.split('/')[-2]
        filename = '%s' % page
        with open(filename, 'wb') as f:
            f.write(response.body)
        self.log('Saved file %s' % filename)

I want to open a webpage and that webpage contains many other links, I want to open all those and wants Scrapy to scrape all those web pages. Please help me out. Thanks in Advance.

Answer 1

I have tried with monsterindia.com and open page using scrapy, that page contain multiple links. I have scraped all the data in the respective link and also we can do pagination. The following code may useful.

    class MonsterSpider(scrapy.Spider):
        name = 'monster'
        start_urls = ['http://jobsearch.monsterindia.com/searchresult.html?day=1&jbc=22']
        item =  BotItem()
        count = 1

        def parse(self, response):
            for href in response.css('h2.seotitle > a::attr(href)'):
                url = response.urljoin(href.extract())
                yield scrapy.Request(url =url, callback = self.parse_details)

            next_page_url = response.css('ul.pager').xpath('//a[contains(text(), "Next")]/@althref').extract_first()
            print next_page_url
            if next_page_url:
               nextpage = response.css('ul.pager').xpath('//a[contains(text(), "Next")]/@onclick').extract_first()
               searchresult_num = nextpage.split("'")[1].strip()
               next_page_url = "http://jobsearch.monsterindia.com/searchresult.html?day=1&n="+searchresult_num
               next_page_url = response.urljoin(next_page_url) 
               print next_page_url
               yield scrapy.Request(url = next_page_url, callback = self.parse)