Python 刮擦问题

我正在尝试使用Scrapy运行刮板，以前我可以使用此代码，但现在我遇到了一个奇怪的错误

 _rules =(Rule(LinkExtractor(restrict_xpaths=(xpath_str)), follow=True,
          callback='parse_url'),)

 def parse_url(self, response):
     print response.url
     ...

基本上，当我运行它时，我得到的是：

Traceback (most recent call last):


 File "/usr/lib/pymodules/python2.7/scrapy/utils/defer.py", line 102, in iter_errback
    yield next(it)
  File "/usr/lib/pymodules/python2.7/scrapy/spidermiddlewares/offsite.py", line 28, in process_spider_output
    for x in result:
  File "/usr/lib/pymodules/python2.7/scrapy/spidermiddlewares/referer.py", line 22, in <genexpr>
    return (_set_referer(r) for r in result or ())
  File "/usr/lib/pymodules/python2.7/scrapy/spidermiddlewares/urllength.py", line 37, in <genexpr>
    return (r for r in result or () if _filter(r))
  File "/usr/lib/pymodules/python2.7/scrapy/spidermiddlewares/depth.py", line 54, in <genexpr>
    return (r for r in result or () if _filter(r))
  File "/usr/lib/pymodules/python2.7/scrapy/spiders/crawl.py", line 67, in _parse_response
    cb_res = callback(response, **cb_kwargs) or ()
TypeError: 'str' object is not callable

---

将

callback='parse\u url'

更改为

callback=self.parse\u url

您是否有一个名为

parse_url

的函数？然后您需要指定

parse\u url

而不是

'parse\u url'

您确定这与规则中的

回调相关吗？请发布您的爬行器的完整代码。@falsetru我有一个函数parse_url，并尝试了您的建议，但它不起作用。@alecxe我非常确定，因为当我删除回调时它确实起作用。该错误表明，在尝试调用回调时，规则类中存在问题。这是完整的代码，非常简单。@ilijaluve首先或全部，为什么是
\u规则
而不是
规则
？另外，检查缩进。
from scrapy.linkextractors import LinkExtractor
from scrapy.spiders import CrawlSpider, Rule
from ..model import Properties


class TestScraper(CrawlSpider):
      name = "test"
      start_urls = [Properties.start_url]

     _rules =( Rule(LinkExtractor(restrict_xpaths=(Properites.xpath)), follow=True, callback='parse_url'), )

        def parse_url(self, response):
            print response.url