我正在尝试爬网一个像YouTube这样的网站,它有一个包含大量视频的列表和一个指向单个视频的链接。我要做的是在使用parse\u item()进入特定视频之前获取视频的缩略图。你知道吗
问题是我不知道如何将“列表视图”的响应对象引入parse\u item()函数。我知道您可以使用process\u request截获请求,并向请求对象插入一个meta,但我不知道如何获得列表视图响应。你知道吗
这个问题有不同的解决方法吗?你知道吗
import re
import datetime
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.selector import Selector
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from ..items import ExampleItem
class ExampleSpider(CrawlSpider):
"""
Crawler for: www.example.com
"""
name = "example"
allowed_domains = ['www.example.com']
start_urls = ['http://www.example.com']
rules = (
Rule(SgmlLinkExtractor(
restrict_xpaths=["//div[@class='pagination']"]
)),
Rule(SgmlLinkExtractor(
restrict_xpaths=["//ul[@class='list']"],
deny=['/user/'],
), callback='parse_item', process_request='parent_url')
)
def parent_url(self, request):
request.meta['parent_page'] = '' # Get the parent response somehow?
return request
def parse_item(self, response):
sel = Selector(response)
item = ExampleItem()
duration = sel.css('.video span::text')[0].extract()
item['title'] = sel.css('.title::text')[0].extract()
item['description'] = sel.xpath('//div[@class="description"]/text()').extract()
item['duration'] = self._parse_duration(duration)
item['link'] = response.url
return item
def _parse_duration(self, string):
"""
Parse the duration field for times
return Datetime object
"""
if len(string) > 20:
return datetime.datetime.strptime(string, '%H hours %M min %S sec').time()
if '60 min' in string:
string.replace('60 min', '01 hours 00 min')
return datetime.datetime.strptime(string, '%H hours %M min %S sec')
return datetime.datetime.strptime(string, '%M min %S sec').time()
我假设您想知道从中提取链接(请求)的URL。你知道吗
您可以重写方法_requests_to_follow,以便传递请求的源页面:
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