Scrapy通过IP地址抓取本地网站

Question

I'm still experimenting with Scrapy, and I'm trying to crawl a website on my local network. The website has the IP address 192.168.0.185. This is my spider:

 from scrapy.spider import BaseSpider
 class 192.168.0.185_Spider(BaseSpider):
      name = "192.168.0.185"
      allowed_domains = ["192.168.0.185"]
      start_urls = ["http://192.168.0.185/"]

      def parse(self, response):
          print "Test:", response.headers

And then in the same directory as my spider I'd execute this shell command to run the spider:

scrapy crawl 192.168.0.185

And I get a very ugly, unreadable error message:

 2012-02-10 20:55:18-0600 [scrapy] INFO: Scrapy 0.14.0 started (bot: tutorial)
 2012-02-10 20:55:18-0600 [scrapy] DEBUG: Enabled extensions: LogStats,   
 TelnetConsole,     CloseSpider, WebService, CoreStats, MemoryUsage, SpiderState
 2012-02-10 20:55:18-0600 [scrapy] DEBUG: Enabled downloader middlewares:      
 HttpAuthMiddleware, DownloadTimeoutMiddleware, UserAgentMiddleware, RetryMiddleware,  
 DefaultHeadersMiddleware, RedirectMiddleware, CookiesMiddleware, 
 HttpCompressionMiddleware, ChunkedTransferMiddleware, DownloaderStats
 2012-02-10 20:55:18-0600 [scrapy] DEBUG: Enabled spider middlewares:   
 HttpErrorMiddleware, OffsiteMiddleware, RefererMiddleware, UrlLengthMiddleware,  
 DepthMiddleware 2012-02-10 20:55:18-0600 [scrapy] DEBUG: Enabled item pipelines: 
 Traceback (most recent call last): File "/usr/bin/scrapy", line 5, in <module>
 pkg_resources.run_script('Scrapy==0.14.0', 'scrapy')
 File "/usr/lib/python2.6/site-packages/pkg_resources.py", line 467, in run_script
 self.require(requires)[0].run_script(script_name, ns)
 File "/usr/lib/python2.6/site-packages/pkg_resources.py", line 1200, in run_script
 execfile(script_filename, namespace, namespace)
 File "/usr/lib/python2.6/site-packages/Scrapy-0.14.0-py2.6.egg/EGG-INFO/scripts   
 /scrapy", line 4, in <module>
 execute()
 File "/usr/lib/python2.6/site-packages/Scrapy-0.14.0-py2.6.egg/scrapy/cmdline.py",  
 line 132, in execute
 _run_print_help(parser, _run_command, cmd, args, opts)
 File "/usr/lib/python2.6/site-packages/Scrapy-0.14.0-py2.6.egg/scrapy/cmdline.py",   
 line 97, in _run_print_help func(*a, **kw)
 File "/usr/lib/python2.6/site-packages/Scrapy-0.14.0-py2.6.egg/scrapy/cmdline.py",  
 line 139, in _run_command cmd.run(args, opts)
 File "/usr/lib/python2.6/site-packages/Scrapy-0.14.0-py2.6.egg/scrapy/commands   
 /crawl.py", line 43, in run
 spider = self.crawler.spiders.create(spname, **opts.spargs)
 File "/usr/lib/python2.6/site-packages/Scrapy-0.14.0-py2.6.egg/scrapy  
 /spidermanager.py", line 43, in create
 raise KeyError("Spider not found: %s" % spider_name)
 KeyError: 'Spider not found: 192.168.0.185'

So then I made another spider, which is practically the same as the first one, except it uses a domain name rather than an IP address. This one worked just fine. Does anyone know what the deal is? How can I get Scrapy to crawl a website via IP address as opposed to a domain name?

from scrapy.spider import BaseSpider
class facebook_Spider(BaseSpider):
     name = "facebook"
     allowed_domains = ["facebook.com"]
     start_urls = ["http://www.facebook.com/"]


     def parse(self, response):
         print "Test:", response.headers

Answer 1

class 192.168.0.185_Spider(BaseSpider):
    ...

You can't use class name which begins with digit or contains dots in Python. See documentation Identifiers and keywords

You can create this spider with correct name:

$ scrapy startproject testproj
$ cd testproj
$ scrapy genspider testspider 192.168.0.185
  Created spider 'testspider' using template 'crawl' in module:
    testproj.spiders.testspider

Spider definition will look like this:

class TestspiderSpider(CrawlSpider):
    name = 'testspider'
    allowed_domains = ['192.168.0.185']
    start_urls = ['http://www.192.168.0.185/']
    ...

And probably you should delete www from start_urls . To start crawling, use spider name instead host:

$ scrapy crawl testspider