Question

这是我的代码：

from scrapy import * 
from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector 
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor

class lala(CrawlSpider):
    name="lala"
    start_url=["http://www.lala.net/"]       
    rule = [Rule(SgmlLinkExtractor(), follow=True, callback='self.parse')] 

    def __init__(self):
        super(lala, self).__init__(self)    
        print "\nworking\n"

    def parse(self,response):        
        print "\n\n Middle \n"  

print "\nend\n"

问题是：

UNFORMATTABLE OBJECT WRITTEN TO LOG with fmt '[%(system)s] %(text)s\n', MESSAGE LOST
2013-04-09 13:48:25+0100 UNFORMATTABLE OBJECT WRITTEN TO LOG with fmt '[%(system)s] %(text)s\n', MESSAGE LOST
2013-04-09 13:48:25+0100 UNFORMATTABLE OBJECT WRITTEN TO LOG with fmt '[%(system)s] %(text)s\n', MESSAGE LOST
2013-04-09 13:48:25+0100 UNFORMATTABLE OBJECT WRITTEN TO LOG with fmt '[%(system)s] %(text)s\n', MESSAGE LOST
2013-04-09 13:48:25+0100 UNFORMATTABLE OBJECT WRITTEN TO LOG with fmt '[%(system)s] %(text)s\n', MESSAGE LOST

请注意，在这种情况下，end和working都会打印出来。

如果我删除了init，那么没有错误但是没有调用解析，因为没有打印中间的msg。

Answer 1

在使用self调用继承的__init__()方法时，

}

super()

查看example listed in the documentation，应该将该属性称为def __init__(self): super(lala, self).__init__()，而不是rules：

rule

Answer 2

scrapy文档explicitly warns against using a CrawlSpider and overridding the parse method。

尝试将parse方法重命名为parse_item，然后重试。

无法格式化的对象错误

2 个答案: