scrapy-解析分页的项目

我有一个形式的网址:

example.com/foo/bar/page_1.html

总共有53页,每页约20行。

我基本上想从所有页面中获取所有行,即〜53 * 20个项目。

我的parse方法中有有效的代码,该代码分析单个页面,每个项目也深入一页,以获取有关该项目的更多信息:

  def parse(self, response):

hxs = HtmlXPathSelector(response)

restaurants = hxs.select('//*[@id="contenido-resbus"]/table/tr[position()>1]')

for rest in restaurants:

item = DegustaItem()

item['name'] = rest.select('td[2]/a/b/text()').extract()[0]

# some items don't have category associated with them

try:

item['category'] = rest.select('td[3]/a/text()').extract()[0]

except:

item['category'] = ''

item['urbanization'] = rest.select('td[4]/a/text()').extract()[0]

# get profile url

rel_url = rest.select('td[2]/a/@href').extract()[0]

# join with base url since profile url is relative

base_url = get_base_url(response)

follow = urljoin_rfc(base_url,rel_url)

request = Request(follow, callback = parse_profile)

request.meta['item'] = item

return request

def parse_profile(self, response):

item = response.meta['item']

# item['address'] = figure out xpath

return item

问题是,如何爬行每个页面?

example.com/foo/bar/page_1.html

example.com/foo/bar/page_2.html

example.com/foo/bar/page_3.html

...

...

...

example.com/foo/bar/page_53.html

回答:

你有两种选择可以解决你的问题。一般的做法是使用yield来生成新请求return。这样,你可以从单个回调中发出多个新请求。

在你的情况下,可能有一个更简单的解决方案:只需从这样的模式中生成启动urs列表:

class MySpider(BaseSpider):

start_urls = ['http://example.com/foo/bar/page_%s.html' % page for page in xrange(1,54)]

以上是 scrapy-解析分页的项目 的全部内容, 来源链接: utcz.com/qa/422140.html

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