Question

我使用python的scrapy工具在python中编写了一个爬虫。以下是python代码：

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
#from scrapy.item import Item
from a11ypi.items import AYpiItem

class AYpiSpider(CrawlSpider):
        name = "AYpi"
        allowed_domains = ["a11y.in"]
        start_urls = ["http://a11y.in/a11ypi/idea/firesafety.html"]

        rules =(
                Rule(SgmlLinkExtractor(allow = ()) ,callback = 'parse_item')
                )

        def parse_item(self,response):
                #filename = response.url.split("/")[-1]
                #open(filename,'wb').write(response.body)
                #testing codes ^ (the above)

                hxs = HtmlXPathSelector(response)
                item = AYpiItem()
                item["foruri"] = hxs.select("//@foruri").extract()
                item["thisurl"] = response.url
                item["thisid"] = hxs.select("//@foruri/../@id").extract()
                item["rec"] = hxs.select("//@foruri/../@rec").extract()
                return item

但是，抛出的错误不是跟随链接而是：

Traceback (most recent call last):
  File "/usr/lib/python2.6/site-packages/Scrapy-0.12.0.2538-py2.6.egg/scrapy/cmdline.py", line 131, in execute
    _run_print_help(parser, _run_command, cmd, args, opts)
  File "/usr/lib/python2.6/site-packages/Scrapy-0.12.0.2538-py2.6.egg/scrapy/cmdline.py", line 97, in _run_print_help
    func(*a, **kw)
  File "/usr/lib/python2.6/site-packages/Scrapy-0.12.0.2538-py2.6.egg/scrapy/cmdline.py", line 138, in _run_command
    cmd.run(args, opts)
  File "/usr/lib/python2.6/site-packages/Scrapy-0.12.0.2538-py2.6.egg/scrapy/commands/crawl.py", line 45, in run
    q.append_spider_name(name, **opts.spargs)
--- <exception caught here> ---
  File "/usr/lib/python2.6/site-packages/Scrapy-0.12.0.2538-py2.6.egg/scrapy/queue.py", line 89, in append_spider_name
    spider = self._spiders.create(name, **spider_kwargs)
  File "/usr/lib/python2.6/site-packages/Scrapy-0.12.0.2538-py2.6.egg/scrapy/spidermanager.py", line 36, in create
    return self._spiders[spider_name](**spider_kwargs)
  File "/usr/lib/python2.6/site-packages/Scrapy-0.12.0.2538-py2.6.egg/scrapy/contrib/spiders/crawl.py", line 38, in __init__
    self._compile_rules()
  File "/usr/lib/python2.6/site-packages/Scrapy-0.12.0.2538-py2.6.egg/scrapy/contrib/spiders/crawl.py", line 82, in _compile_rules
    self._rules = [copy.copy(r) for r in self.rules]
exceptions.TypeError: 'Rule' object is not iterable

有人可以向我解释发生了什么事吗？由于这是文档中提到的内容，我将允许字段留空，因此默认情况下应该遵循True。那么为什么会出错呢？我可以使用爬虫来快速进行哪种优化？

Answer 1

从我看来，看起来你的规则不是可迭代的。看起来你试图将规则作为一个元组，你应该read up on tuples in the python documentation。

要解决您的问题，请更改此行：

    rules =(
            Rule(SgmlLinkExtractor(allow = ()) ,callback = 'parse_item')
            )

要：

    rules =(Rule(SgmlLinkExtractor(allow = ()) ,callback = 'parse_item'),)

注意最后的逗号？

python中的scrapy Crawler无法跟踪链接？

1 个答案: