python - 使用 Scrapy 在 Craigslist 上进行递归抓取

Question

我一直在尝试通过构建 scrapers 来磨练我的 python 技能，最近从 bs4 切换到 scrapy，以便我可以使用它的多线程和下载延迟功能。我已经能够制作一个基本的抓取器并将数据输出到 csv，但是当我尝试添加递归功能时，我遇到了问题。我尝试遵循 Scrapy Recursive download of Content 的建议但不断收到以下错误:

调试:重试 http://medford.craigslist.org%20%5Bu'/cto/4359874426.html'%5D> DNS 查找失败:找不到地址

这让我觉得我尝试加入链接的方式不起作用，因为它将字符插入到网址中，但我不知道如何修复它。有什么建议吗？

代码如下:

#-------------------------------------------------------------------------------
# Name:        module1
# Purpose:
#
# Author:      CD
#
# Created:     02/03/2014
# Copyright:   (c) CD 2014
# Licence:     <your licence>
#-------------------------------------------------------------------------------
from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector
from craigslist_sample.items import CraigslistSampleItem
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.http import Request
from scrapy.selector import *

class PageSpider(BaseSpider):
    name = "cto"

    start_urls = ["http://medford.craigslist.org/cto/"]

    rules = (Rule(SgmlLinkExtractor(allow=("index\d00\.html", ), restrict_xpaths=('//p[@class="nextpage"]' ,))
        , callback="parse", follow=True), )

    def parse(self, response):
        hxs = HtmlXPathSelector(response)
        titles = hxs.select("//span[@class='pl']")
        for titles in titles:
            item = CraigslistSampleItem()
            item['title'] = titles.select("a/text()").extract()
            item['link'] = titles.select("a/@href").extract()

            url = "http://medford.craiglist.org %s" % item['link']
            yield Request(url=url, meta={'item': item}, callback=self.parse_item_page)

    def parse_item_page(self, response):
        hxs = HtmlXPathSelector(response)

        item = response.meta['item']
        item['description'] = hxs.select('//section[@id="postingbody"]/text()').extract()
        return item

Answer 1

结果是你的代码:

 url = "http://medford.craiglist.org %s" % item['link']

生成:

http://medford.craigslist.org [u'/cto/4359874426.html']

item['link'] 返回代码中的列表，而不是您期望的字符串。您需要这样做:

url = 'http://medford.craiglist.org{}'.format(''.join(item['link']))