#我的代码如下,问题在代码的注释里面
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor as sle #用于定义需要提取的链接
class NewsSpider(CrawlSpider):
<code>name = "demo2" allowed_domains = ["360.cn"] # start_urls=["http://bobao.360.cn/activity/index&page=2"]#这个连接测试成功 # rules = [ Rule(sle(allow=r'/vul/index?type=all&page=\d{1,3}'), follow=True,callback='parse_item1') ]#*这个测试不成功原因好像* #allow=r'/vul/index?type=all&page=\d{1,3} 这个里面/vul/index?这个?有影响,请教如何修改 rules = [ Rule(sle(allow=r'/activity/index&page=\d{1,3}'), follow=True,callback='parse_item1') ]#这个连接测试成功 </code>
<code>def parse_item1(self, response): print u'这是谁?????????????????????' </code>
#我的代码如下,问题在代码的注释里面
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor as sle #用于定义需要提取的链接
class NewsSpider(CrawlSpider):
<code>name = "demo2" allowed_domains = ["360.cn"] # start_urls=["http://bobao.360.cn/activity/index&page=2"]#这个连接测试成功 # rules = [ Rule(sle(allow=r'/vul/index?type=all&page=\d{1,3}'), follow=True,callback='parse_item1') ]#*这个测试不成功原因好像* #allow=r'/vul/index?type=all&page=\d{1,3} 这个里面/vul/index?这个?有影响,请教如何修改 rules = [ Rule(sle(allow=r'/activity/index&page=\d{1,3}'), follow=True,callback='parse_item1') ]#这个连接测试成功 </code>
<code>def parse_item1(self, response): print u'这是谁?????????????????????' </code>
网页中链接是这样:#/rwxwsblog/default.html?page=3"
要写成这样的:Rule(sle(allow=("/rwxwsblog/default.html\?page=\d{1,}")), #此处要注意?号的转换,复制过来需要对?号进行转义。