Question

我一直试图通过构建刮刀来磨练我的python技能，最近从bs4切换到scrapy，这样我就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刮刀并将数据输出到csv，但是当我尝试添加递归功能时，我遇到了问题。我尝试按照Scrapy Recursive download of Content的建议，但一直收到以下错误：

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

这让我觉得我试图加入链接的方式不起作用，因为它将字符插入到网址中，但我无法弄清楚如何修复它。有什么建议吗？

以下是代码：

#-------------------------------------------------------------------------------
# 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

结果你的代码：

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

产生

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

item['link']会在您的代码中返回一个列表，而不是您期望的字符串。你需要这样做：

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

用Scrapy进行Craigslist上的递归刮擦

1 个答案: