Question

我正在尝试：

class SpiderSpider(CrawlSpider):
    name = "lolies"
    allowed_domains = ["domain.com"]
    start_urls = ['http://www.domain.com/directory/lol2']
    rules = (Rule(SgmlLinkExtractor(allow=[r'directory/lol2/\w+$']), follow=True), Rule(SgmlLinkExtractor(allow=[r'directory/lol2/\w+/\d+$']), follow=True),Rule(SgmlLinkExtractor(allow=[r'directory/lol2/\d+$']), callback=self.parse_loly))

def parse_loly(self, response):
    print 'Hi this is the loly page %s' % response.url
    return

那让我回头：

NameError: name 'self' is not defined

如果我将回调更改为callback="self.parse_loly"似乎永远不会被调用并打印该URL。

但似乎是在没有问题的情况下抓取网站，因为我为该规则获得了许多Crawled 200调试消息。

我可能做错了什么？

先谢谢你们！

Answer 1

看起来parse_loly的空格未正确对齐。 Python对空格敏感，因此对于解释器来说，它看起来像是SpiderSpider之外的方法。

您可能还希望按照PEP8将规则行拆分为较短的行。

试试这个：

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor

class SpiderSpider(CrawlSpider):
    name = "lolies"
    allowed_domains = ["domain.com"]
    start_urls = ['http://www.domain.com/directory/lol2/']
    rules = (
        Rule(SgmlLinkExtractor(allow=('\w+$', ))), 
        Rule(SgmlLinkExtractor(allow=('\w+/\d+$', ))),
        Rule(SgmlLinkExtractor(allow=('\d+$',)), callback='parse_loly'),
    )

    def parse_loly(self, response):
        print 'Hi this is the loly page %s' % response.url
        return None

Scrapy SgmlLinkExtractor规则和回调令人头疼

1 个答案: