Python 抓取图像的url

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

---

您正在提取链接（“a”元素），而不是图像（“img”元素）。试试这个：

# 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

---

寻求调试帮助的问题（“为什么这段代码不起作用？”）必须包括所需的行为、特定的问题或错误以及在问题本身中重现它所需的最短代码。没有明确问题陈述的问题对其他读者没有用处。请参阅：。