抓取图像的网址

Question

如何使用python中的scrapy从网站获取图像URL。请帮助我。这是我的代码

from scrapy.spiders import CrawlSpider, Rule
#from scrapy.linkextractors.lxmlhtml import LxmlLinkExtractor
from scrapy.contrib.linkextractors import LinkExtractor
from scrapy.item import Item, Field

class MyItem(Item):
    url= Field()


class someSpider(CrawlSpider):
    name = 'crawltest'
    allowed_domains = ['bambeeq.com']
    start_urls = ['http://www.bambeeq.com/']
    rules = (Rule(LinkExtractor(allow=()), callback='parse_obj', follow=True),)

    def parse_obj(self,response):
        item = MyItem()
        item['url'] = []
        for link in LinkExtractor(allow=(),deny = self.allowed_domains).extract_links(response):
            item['url'].append(link.url)
            #item['image'].append(link.img)
        return item

Answer 1

您正在提取链接（“ a”元素），而不是图像（“ img”元素）。 尝试这个：

# iterate over the list of images
for image in response.xpath('//img/@src').extract():
    # make each one into a full URL and add to item[]
    item['url'].append(response.urljoin(image))

yield item