Scrapy python 错误 - 请求中缺少方案 URL

Question

我正在尝试从受密码保护的 FTP 服务器中提取文件。这是我正在使用的代码：

import scrapy
from scrapy.contrib.spiders import XMLFeedSpider
from scrapy.http import Request
from crawler.items import CrawlerItem

class SiteSpider(XMLFeedSpider):
    name = 'site'
    allowed_domains = ['ftp.site.co.uk']
    itertag = 'item'

    def start_requests(self):
        yield Request('ftp.site.co.uk/feed.xml',
            meta={'ftp_user': 'test', 'ftp_password': 'test'})

    def parse_node(self, response, selector):
        item = CrawlerItem()
        item['title'] = (selector.xpath('//title/text()').extract() or [''])[0]      
        return item

这是我得到的回溯错误：

        Traceback (most recent call last):                                                              
          File "/usr/local/lib/python2.7/dist-packages/twisted/internet/base.py", line 1192, in run     
            self.mainLoop()                                                                             
          File "/usr/local/lib/python2.7/dist-packages/twisted/internet/base.py", line 1201, in mainLoop
            self.runUntilCurrent()                                                                      
          File "/usr/local/lib/python2.7/dist-packages/twisted/internet/base.py", line 824, in runUntilC
urrent                                                                                                  
            call.func(*call.args, **call.kw)                                                            
          File "/usr/local/lib/python2.7/dist-packages/scrapy/utils/reactor.py", line 41, in __call__   
            return self._func(*self._a, **self._kw)                                                     
        --- <exception caught here> ---                                                                 
          File "/usr/local/lib/python2.7/dist-packages/scrapy/core/engine.py", line 112, in _next_reques
t                                                                                                       
            request = next(slot.start_requests)                                                         
          File "/var/www/spider/crawler/spiders/site.py", line 13, in start_requests                 
            meta={'ftp_user': 'test', 'ftp_password': 'test'})                                
          File "/usr/local/lib/python2.7/dist-packages/scrapy/http/request/__init__.py", line 26, in __i
nit__                                                                                                   
            self._set_url(url)                                                                          
          File "/usr/local/lib/python2.7/dist-packages/scrapy/http/request/__init__.py", line 61, in _se
t_url                                                                                                   
            raise ValueError('Missing scheme in request url: %s' % self._url)                           
        exceptions.ValueError: Missing scheme in request url: ftp.site.co.uk/f
eed.xml

Answer 1

您需要为 URL 添加方案：

ftp://ftp.site.co.uk

FTP URL 语法定义为：

ftp://[<user>[:<password>]@]<host>[:<port>]/<url-path>

基本上，你这样做：

yield Request('ftp://ftp.site.co.uk/feed.xml', ...)

在维基百科上阅读更多关于架构的信息：http://en.wikipedia.org/wiki/URI_scheme

Scrapy python 错误 - 请求中缺少方案 URL

Scrapy python error - Missing scheme in request URL

python

web-crawler

scrapy

scrapy-spider