Scrapy start_urls

Question

The script（下）来自this 教程包含两个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

但是为什么它只抓取这两个网页呢？我看到allowed_domains = ["dmoz.org"] 但这两个页面还包含指向dmoz.org 域内的其他页面的链接！为什么它也不刮掉它们？

Answer 1

start_urls 类属性包含起始 url - 仅此而已。如果您已经提取了您想要抓取的其他页面的 url - yield from parse 回调对应的请求与 [another] 回调：

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):
        ...

如果您仍想自定义启动请求创建，请覆盖方法BaseSpider.start_requests()

Answer 2

start_urls 包含蜘蛛开始爬行的那些链接。 如果你想递归爬行，你应该使用 crawlspider 并为此定义规则。 http://doc.scrapy.org/en/latest/topics/spiders.html 看看那里的例子。

Answer 3

该类没有rules 属性。查看http://readthedocs.org/docs/scrapy/en/latest/intro/overview.html 并搜索“规则”以查找示例。

Answer 4

如果你使用BaseSpider，在回调中，你必须自己提取你想要的url并返回一个Request对象。

如果您使用CrawlSpider，链接提取将由规则和与规则关联的 SgmlLinkExtractor 负责。

Answer 5

如果你使用规则来跟踪链接（已经在 scrapy 中实现），蜘蛛也会抓取它们。希望对您有所帮助...

    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

你没有写函数来处理你想要得到的url。所以有两种方法来reslolve。1.使用规则（crawlspider）2：编写处理新url的函数。并将它们放入回调函数。