Python 3.x 如何在python中提取网页中的src元素
我需要从'https://www.gizbot.com/mobile-brands-in-india/'. 我试过用scrapy做这件事- spider.pyPython 3.x 如何在python中提取网页中的src元素,python-3.x,web-scraping,scrapy,Python 3.x,Web Scraping,Scrapy,我需要从'https://www.gizbot.com/mobile-brands-in-india/'. 我试过用scrapy做这件事- spider.py def parse(self, response): page = response.url.split("/")[-2] filename = 'mobiles-%s.html' % page mob = response.xpath('.//div[has-c
def parse(self, response):
page = response.url.split("/")[-2]
filename = 'mobiles-%s.html' % page
mob = response.xpath('.//div[has-class("all-brands-block-desc-brand")]/text()').getall()
for mobile in mob:
m = str(mobile).split()[0]
with open(filename, 'a') as f:
f.write("%s %s\n" % (mobile, response.xpath('.//a[contains(@href, m)]').xpath("@href").extract()))
self.log('Saved file %s' % filename)
但是它没有提取正确的数据。我不知道哪里出了问题。非常感谢您的帮助。您需要使用以下xpath:
mob = response.xpath('//div[contains(@class, "all-brands-block-desc-brand")]').getall()
xpath*对于提取图像src,您有一个输入错误。对于i in response.css('div.all-brands-block'):print(“+i.css('img::attr('data pagespeed lazy src')))。get())