Scrapy start_urls

Question

The script (below) from this tutorial contains two start_urls .

from scrapy.spider import Spider
from scrapy.selector import Selector

from dirbot.items import Website

class DmozSpider(Spider):
    name = "dmoz"
    allowed_domains = ["dmoz.org"]
    start_urls = [
        "http://www.dmoz.org/Computers/Programming/Languages/Python/Books/",
        "http://www.dmoz.org/Computers/Programming/Languages/Python/Resources/",
    ]

    def parse(self, response):
        """
        The lines below is a spider contract. For more info see:
        http://doc.scrapy.org/en/latest/topics/contracts.html
        @url http://www.dmoz.org/Computers/Programming/Languages/Python/Resources/
        @scrapes name
        """
        sel = Selector(response)
        sites = sel.xpath('//ul[@class="directory-url"]/li')
        items = []

        for site in sites:
            item = Website()
            item['name'] = site.xpath('a/text()').extract()
            item['url'] = site.xpath('a/@href').extract()
            item['description'] = site.xpath('text()').re('-\s[^\n]*\\r')
            items.append(item)

        return items

But why does it scrape only these 2 web pages? I see allowed_domains = ["dmoz.org"] but these two pages also contain links to other pages which are within dmoz.org domain! Why doesnt it scrape them too?

Answer 1

start_urls class attribute contains start urls - nothing more. If you have extracted urls of other pages you want to scrape - yield from parse callback corresponding requests with [another] callback:

class Spider(BaseSpider):

    name = 'my_spider'
    start_urls = [
                'http://www.domain.com/'
    ]
    allowed_domains = ['domain.com']

    def parse(self, response):
        '''Parse main page and extract categories links.'''
        hxs = HtmlXPathSelector(response)
        urls = hxs.select("//*[@id='tSubmenuContent']/a[position()>1]/@href").extract()
        for url in urls:
            url = urlparse.urljoin(response.url, url)
            self.log('Found category url: %s' % url)
            yield Request(url, callback = self.parseCategory)

    def parseCategory(self, response):
        '''Parse category page and extract links of the items.'''
        hxs = HtmlXPathSelector(response)
        links = hxs.select("//*[@id='_list']//td[@class='tListDesc']/a/@href").extract()
        for link in links:
            itemLink = urlparse.urljoin(response.url, link)
            self.log('Found item link: %s' % itemLink, log.DEBUG)
            yield Request(itemLink, callback = self.parseItem)

    def parseItem(self, response):
        ...

If you still want to customize start requests creation, override method BaseSpider.start_requests()

Answer 2

start_urls contain those links from which the spider start crawling. If you want crawl recursively you should use crawlspider and define rules for that. http://doc.scrapy.org/en/latest/topics/spiders.html look there for example.

Answer 3

The class does not have a rules property. Have a look at http://readthedocs.org/docs/scrapy/en/latest/intro/overview.html and search for "rules" to find an example.

Answer 4

If you use BaseSpider , inside the callback, you have to extract out your desired urls yourself and return a Request object.

If you use CrawlSpider , links extraction would be taken care of by the rules and the SgmlLinkExtractor associated with the rules.

Answer 5

If you use an rule to follow links (that is already implemented in scrapy), the spider will scrape them too. I hope have helped...

    from scrapy.contrib.spiders import BaseSpider, Rule
    from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
    from scrapy.selector import HtmlXPathSelector


    class Spider(BaseSpider):
        name = 'my_spider'
        start_urls = ['http://www.domain.com/']
        allowed_domains = ['domain.com']
        rules = [Rule(SgmlLinkExtractor(allow=[], deny[]), follow=True)]

     ...

Answer 6

你没有编写函数来处理url你想要得到的东西。还有两种方法来解析。使用规则（crawlspider）2：编写处理新urls的函数。并将它们放在回调函数中。