Scrapy通过IP地址爬取本地网站

Question

我正在尝试使用Scrapy这个工具，想要爬取我本地网络上的一个网站。这个网站的IP地址是192.168.0.185。以下是我的爬虫代码：

 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

然后在和我的爬虫代码同一个文件夹里，我会执行这个命令来运行爬虫：

scrapy crawl 192.168.0.185

但是我收到了一个非常难看、难以理解的错误信息：

 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'

于是我又做了一个爬虫，几乎和第一个一模一样，只是这次用的是域名，而不是IP地址。这个爬虫运行得很好。有没有人知道这是怎么回事？我该如何让Scrapy通过IP地址而不是域名来爬取网站呢？

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):
    ...

在Python中，类名不能以数字开头，也不能包含点（.）。你可以查看文档了解更多信息，链接在这里：标识符和关键字

你可以用正确的名字来创建这个爬虫：

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

爬虫的定义看起来是这样的：

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

而且你可能需要从start_urls中删除www。要开始爬取数据，使用爬虫的名字代替主机名：

$ scrapy crawl testspider