如何用Scrapy爬取整个网站？

Question

我无法抓取整个网站，Scrapy 只能抓取表面内容，我想要更深入地抓取。我已经在网上搜索了5到6个小时，但没有找到解决办法。下面是我的代码：

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from scrapy.item import Item
from scrapy.spider import BaseSpider
from scrapy import log

class ExampleSpider(CrawlSpider):
    name = "example.com"
    allowed_domains = ["example.com"]
    start_urls = ["http://www.example.com/"]
    rules = [Rule(SgmlLinkExtractor(allow=()), 
                  follow=True),
             Rule(SgmlLinkExtractor(allow=()), callback='parse_item')
    ]
    def parse_item(self,response):
        self.log('A response from %s just arrived!' % response.url)

Answer 1

在解析start_urls的时候，可以通过href这个标签来找到更深层次的网址。然后，可以在parse()这个函数里发出更深层的请求。这里有一个简单的例子。下面是最重要的源代码：

from scrapy.spiders import Spider
from tutsplus.items import TutsplusItem
from scrapy.http    import Request
import re

class MySpider(Spider):
    name            = "tutsplus"
    allowed_domains = ["code.tutsplus.com"]
    start_urls      = ["http://code.tutsplus.com/"]

    def parse(self, response):
        links = response.xpath('//a/@href').extract()

        # We stored already crawled links in this list
        crawledLinks = []

        # Pattern to check proper link
        # I only want to get tutorial posts
        linkPattern = re.compile("^\/tutorials\?page=\d+")

        for link in links:
        # If it is a proper link and is not checked yet, yield it to the Spider
            if linkPattern.match(link) and not link in crawledLinks:
                link = "http://code.tutsplus.com" + link
                crawledLinks.append(link)
                yield Request(link, self.parse)

        titles = response.xpath('//a[contains(@class, "posts__post-title")]/h1/text()').extract()
        for title in titles:
            item = TutsplusItem()
            item["title"] = title
            yield item

Answer 2

规则是短路的，这意味着一个链接满足的第一个规则就是会被应用的规则，你的第二个规则（带有回调函数的）将不会被调用。

把你的规则改成这样：

rules = [Rule(SgmlLinkExtractor(), callback='parse_item', follow=True)]