Scrapy 404错误:未处理或不允许HTTP状态代码
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我正在尝试使用scrapy刮取网站coursetalk,我正在尝试使用蜘蛛模板并获得404错误:
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:
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
来自scrapy的竞争日志:
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)
答案
看起来这个网站是如此奇怪,响应状态代码是404但仍然可以正常获取正文。
在Scrapy中,HttpErrorMiddleware默认启用,这将过滤掉不成功的Http响应,以便蜘蛛不必处理它们。在这种情况下,scrapy提供HTTPERROR_ALLOWED_CODES
设置,即使返回错误代码也允许处理响应。
在HTTPERROR_ALLOWED_CODES =[404]
项目中添加setting.py
可以绕过这个问题
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)
另一答案
我已经用scrapy面对这个问题并解决了它。
在USER_AGENT
改变了setting.py
USER_AGENT = "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (Khtml, like Gecko) Chrome/34.0.1847.131 Safari/537.36"
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