如何关注此示例中的链接:http://snippets.scrapy.org/snippets/7/? 访问第一页的链接后脚本停止。
class MySpider(BaseSpider):
"""Our ad-hoc spider"""
name = "myspider"
start_urls = ["http://stackoverflow.com/"]
question_list_xpath = '//div[@id="content"]//div[contains(@class, "question-summary")]'
def parse(self, response):
hxs = HtmlXPathSelector(response)
for qxs in hxs.select(self.question_list_xpath):
loader = XPathItemLoader(QuestionItem(), selector=qxs)
loader.add_xpath('title', './/h3/a/text()')
loader.add_xpath('summary', './/h3/a/@title')
loader.add_xpath('tags', './/a[@rel="tag"]/text()')
loader.add_xpath('user', './/div[@class="started"]/a[2]/text()')
loader.add_xpath('posted', './/div[@class="started"]/a[1]/span/@title')
loader.add_xpath('votes', './/div[@class="votes"]/div[1]/text()')
loader.add_xpath('answers', './/div[contains(@class, "answered")]/div[1]/text()')
loader.add_xpath('views', './/div[@class="views"]/div[1]/text()')
yield loader.load_item()
我试图改变:
class MySpider(BaseSpider):
要
class MySpider(CrawlSpider)
并添加
rules = (
Rule(SgmlLinkExtractor(allow=()),
callback='parse',follow=True),
)
但它不会抓取所有网站
谢谢,
答案 0 :(得分:0)
是的,您需要继承CrawlSpider,并将parse
函数重命名为parse_page
,因为CrawlSpider使用parse
开始抓取。
This was already answered