Question

我希望你们在健康和研发工作方面做得最好。

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)

我想打开一个网页，该网页包含许多其他链接，我想打开所有这些，并希望Scrapy废弃所有这些网页。请帮帮我。提前致谢。

Answer 1

我已尝试使用monsterindia.com并使用scrapy打开页面，该页面包含多个链接。我已经删除了相应链接中的所有数据，我们也可以进行分页。以下代码可能有用。

    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)

如何在网页上打开多个链接并废弃数据？

1 个答案: