Scrapy 抓取提取的链接

Question

我需要抓取一个网站，并在特定的 xpath 上抓取该网站的每个 url 例如。：我需要抓取“http://someurl.com/world/" which has 10 links in the container (xpath("//div[@class='pane-content']")) and i need to crawl all those 10 links and extract images from them, but the links in "http://someurl.com/world/”的样子 “http://someurl.com/node/xxxx”

我目前拥有的：

import scrapy
from scrapy.contrib.spiders import Rule, CrawlSpider
from scrapy.contrib.linkextractors import LinkExtractor
from imgur.items import ImgurItem

class ImgurSpider(CrawlSpider):
    name = 'imgur'
    allowed_domains = ['someurl.com/']
    start_urls = ['http://someurl.com/news']
    rules = [Rule(LinkExtractor(allow=('/node/.*')), callback='parse_imgur', follow=True)]

    def parse_imgur(self, response):
        image = ImgurItem()
        image['title'] = response.xpath(\
            "//h1[@class='pane-content']/a/text()").extract()
        rel = response.xpath("//img/@src").extract()
        image['image_urls'] = response.xpath("//img/@src").extract()
        return image

Answer 1

您可以重写 'Rule' 以满足您的所有要求：

rules = [Rule(LinkExtractor(allow=('/node/.*',), restrict_xpaths=('//div[@class="pane-content"]',)), callback='parse_imgur', follow=True)]

要从提取的图片链接中下载图片，您可以使用 Scrapy 的捆绑包 ImagePipeline

Scrapy 抓取提取的链接

Scrapy crawl extracted links

python

hyperlink

scrapy

extractor