Scrapy 404 错误:未处理或不允许 HTTP 状态代码 [英] Scrapy 404 error: HTTP status code is not handled or not allowed

查看:22
本文介绍了Scrapy 404 错误:未处理或不允许 HTTP 状态代码的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在尝试使用 scrapy 抓取网站 coursetalk,我首先尝试使用蜘蛛模板并收到 404 错误:

I'm trying to scrape the site coursetalk using scrapy, I'm trying with the spider template first and getting a 404 error:

2017-12-29 23:34:30 [scrapy] DEBUG: Ignoring response <404 https://www.coursetalk.com/subjects/data-science/courses/>: HTTP status code is not handled or not allowed

这是我正在使用的代码:

This is the code I'm using:

import scrapy

class ListaDeCursosSpider(scrapy.Spider):
    name = "lista_de_cursos"
    start_urls = ['https://www.coursetalk.com/subjects/data-science/courses/'] 


    def parse(self, response):
        print response.body

以及来自scrapy的竞争日志:

And the compete log from scrapy:

2017-12-29 23:34:26 [scrapy] INFO: Scrapy 1.0.3 started (bot: coursetalk)
2017-12-29 23:34:26 [scrapy] INFO: Optional features available: ssl, http11, boto
2017-12-29 23:34:26 [scrapy] INFO: Overridden settings: {'NEWSPIDER_MODULE': 'coursetalk.spiders', 'SPIDER_MODULES': ['coursetalk.spiders'], 'BOT_NAME': 'coursetalk'}
2017-12-29 23:34:27 [scrapy] INFO: Enabled extensions: CloseSpider, TelnetConsole, LogStats, CoreStats, SpiderState
2017-12-29 23:34:27 [boto] DEBUG: Retrieving credentials from metadata server.
2017-12-29 23:34:28 [boto] ERROR: Caught exception reading instance data
Traceback (most recent call last):
  File "/usr/local/lib/python2.7/dist-packages/boto/utils.py", line 210, in retry_url
    r = opener.open(req, timeout=timeout)
  File "/usr/lib/python2.7/urllib2.py", line 404, in open
    response = self._open(req, data)
  File "/usr/lib/python2.7/urllib2.py", line 422, in _open
    '_open', req)
  File "/usr/lib/python2.7/urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 1214, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/lib/python2.7/urllib2.py", line 1184, in do_open
    raise URLError(err)
URLError: <urlopen error timed out>
2017-12-29 23:34:28 [boto] ERROR: Unable to read instance data, giving up
2017-12-29 23:34:28 [scrapy] INFO: Enabled downloader middlewares: HttpAuthMiddleware, DownloadTimeoutMiddleware, UserAgentMiddleware, RetryMiddleware, DefaultHeadersMiddleware, MetaRefreshMiddleware, HttpCompressionMiddleware, RedirectMiddleware, CookiesMiddleware, ChunkedTransferMiddleware, DownloaderStats
2017-12-29 23:34:28 [scrapy] INFO: Enabled spider middlewares: HttpErrorMiddleware, OffsiteMiddleware, RefererMiddleware, UrlLengthMiddleware, DepthMiddleware
2017-12-29 23:34:28 [scrapy] INFO: Enabled item pipelines: 
2017-12-29 23:34:28 [scrapy] INFO: Spider opened
2017-12-29 23:34:28 [scrapy] INFO: Crawled 0 pages (at 0 pages/min), scraped 0 items (at 0 items/min)
2017-12-29 23:34:28 [scrapy] DEBUG: Telnet console listening on 127.0.0.1:6023
2017-12-29 23:34:30 [scrapy] DEBUG: Crawled (404) <GET https://www.coursetalk.com/subjects/data-science/courses/> (referer: None)
2017-12-29 23:34:30 [scrapy] DEBUG: Ignoring response <404 https://www.coursetalk.com/subjects/data-science/courses/>: HTTP status code is not handled or not allowed
2017-12-29 23:34:30 [scrapy] INFO: Closing spider (finished)

推荐答案

看起来这个网站很奇怪,响应状态码是 404 但仍然可以正常获取正文.

Looks like this website is so weird that response status code is 404 but still can fetch the body normally.

在 Scrapy 中 HttpErrorMiddleware 是默认启用的,它会过滤掉不成功的 Http 响应,这样蜘蛛就不必处理它们.在这种情况下,scrapy 提供了 HTTPERROR_ALLOWED_CODES 设置以允许处理响应,即使返回错误代码.

And in Scrapy HttpErrorMiddleware is default enabled ,which would filter out unsuccessful Http responses so that spiders don't have to deal with them.And in this case,scrapy provides HTTPERROR_ALLOWED_CODES setting to allows to deal with response even if returning error codes.

并且在项目setting.py中添加HTTPERROR_ALLOWED_CODES =[404]将绕过这个问题

And adding HTTPERROR_ALLOWED_CODES =[404] in the project setting.py would bypass this issue

import scrapy
import logging

class ListaDeCursosSpider(scrapy.Spider):
    name = "lista_de_cursos"
    allowed_domains = ['www.coursetalk.com']
    start_urls = ['https://www.coursetalk.com/subjects/data-science/courses/'] 

 def parse(self, response):
        logging.info("response.status:%s"%response.status)
        logourl = response.selector.css('div.main-nav__logo img').xpath('@src').extract()
        logging.info('response.logourl:%s'%logourl)

这篇关于Scrapy 404 错误:未处理或不允许 HTTP 状态代码的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆