使Scrapy跟踪链接并收集数据
我试图用Scrapy编写程序以打开链接并从此标签收集数据:<p class="attrgroup"></p>
。
我设法使Scrapy从给定的URL收集了所有链接,但没有关注它们。任何帮助都非常感谢。
回答:
你需要Request为链接提供实例,分配回调并在回调中提取所需p
元素的文本:
# -*- coding: utf-8 -*-import scrapy
# item class included here
class DmozItem(scrapy.Item):
# define the fields for your item here like:
link = scrapy.Field()
attr = scrapy.Field()
class DmozSpider(scrapy.Spider):
name = "dmoz"
allowed_domains = ["craigslist.org"]
start_urls = [
"http://chicago.craigslist.org/search/emd?"
]
BASE_URL = 'http://chicago.craigslist.org/'
def parse(self, response):
links = response.xpath('//a[@class="hdrlnk"]/@href').extract()
for link in links:
absolute_url = self.BASE_URL + link
yield scrapy.Request(absolute_url, callback=self.parse_attr)
def parse_attr(self, response):
item = DmozItem()
item["link"] = response.url
item["attr"] = "".join(response.xpath("//p[@class='attrgroup']//text()").extract())
return item
以上是 使Scrapy跟踪链接并收集数据 的全部内容, 来源链接: utcz.com/qa/409083.html