Question

我无法抓取整个网站，Scrapy只是在地面爬行，我想爬得更深。谷歌搜索过去5-6小时，没有任何帮助。我的代码如下：

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

规则短路，意味着链接满足的第一条规则将是应用的规则，不会调用第二条规则（带回调）。

将规则更改为：

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

Answer 2

解析start_urls时，标签href可以解析更深层的网址。然后，可以在函数parse()中产生更深的请求。 Here is a simple example。最重要的源代码如下所示：

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

如何使用Scrapy抓取整个网站？

2 个答案: