python - 为什么 scrapy 在尝试抓取和解析网站时会为我抛出错误？

Question

下面的代码

class SiteSpider(BaseSpider):
    name = "some_site.com"
    allowed_domains = ["some_site.com"]
    start_urls = [
        "some_site.com/something/another/PRODUCT-CATEGORY1_10652_-1__85667",
    ]
    rules = (
        Rule(SgmlLinkExtractor(allow=('some_site.com/something/another/PRODUCT-CATEGORY_(.*)', ))),

        # Extract links matching 'item.php' and parse them with the spider's method parse_item
        Rule(SgmlLinkExtractor(allow=('some_site.com/something/another/PRODUCT-DETAIL(.*)', )), callback="parse_item"),
    )
    def parse_item(self, response):
.... parse stuff

抛出以下错误

Traceback (most recent call last):
  File "/usr/lib/python2.6/dist-packages/twisted/internet/base.py", line 1174, in mainLoop
    self.runUntilCurrent()
  File "/usr/lib/python2.6/dist-packages/twisted/internet/base.py", line 796, in runUntilCurrent
    call.func(*call.args, **call.kw)
  File "/usr/lib/python2.6/dist-packages/twisted/internet/defer.py", line 318, in callback
    self._startRunCallbacks(result)
  File "/usr/lib/python2.6/dist-packages/twisted/internet/defer.py", line 424, in _startRunCallbacks
    self._runCallbacks()
--- <exception caught here> ---
  File "/usr/lib/python2.6/dist-packages/twisted/internet/defer.py", line 441, in _runCallbacks
    self.result = callback(self.result, *args, **kw)
  File "/usr/lib/pymodules/python2.6/scrapy/spider.py", line 62, in parse
    raise NotImplementedError
exceptions.NotImplementedError: 

当我将回调更改为“parse”并将函数更改为“parse”时，我没有收到任何错误，但没有任何内容被删除。我将其更改为“parse_items”，认为我可能会覆盖 parse method by accident .也许我错误地设置了链接提取器？

我想做的是解析 CATEGORY 页面上的每个 ITEM 链接。我这样做完全错了吗？

Answer 1

我需要将 BaseSpider 更改为 CrawlSpider。感谢 srapy 用户!

http://groups.google.com/group/scrapy-users/browse_thread/thread/4adaba51f7bcd0af#

Hi Bob,

Perhaps it might work if you change from BaseSpider to CrawlSpider? The BaseSpider seems not implement Rule, see:

http://doc.scrapy.org/topics/spiders.html?highlight=rule#scrapy.contr...

-M