Scrapy CrawlSpider不会抓取第一个登录页面

Question

我是Scrapy的新手，正在从事刮擦练习，并且正在使用CrawlSpider。尽管Scrapy框架运行良好并且遵循相关链接，但我似乎无法使CrawlSpider抓取第一个链接（主页/登录页面）。相反，它直接刮取规则确定的链接，但不刮取链接所在的登录页面。我不知道如何解决此问题，因为不建议覆盖CrawlSpider的parse方法。修改follow
= True / False也不会产生任何好的结果。这是代码片段：

class DownloadSpider(CrawlSpider):
    name = 'downloader'
    allowed_domains = ['bnt-chemicals.de']
    start_urls = [
        "http://www.bnt-chemicals.de"        
        ]
    rules = (   
        Rule(SgmlLinkExtractor(aloow='prod'), callback='parse_item', follow=True),
        )
    fname = 1

    def parse_item(self, response):
        open(str(self.fname)+ '.txt', 'a').write(response.url)
        open(str(self.fname)+ '.txt', 'a').write(','+ str(response.meta['depth']))
        open(str(self.fname)+ '.txt', 'a').write('\n')
        open(str(self.fname)+ '.txt', 'a').write(response.body)
        open(str(self.fname)+ '.txt', 'a').write('\n')
        self.fname = self.fname + 1

Answer 1

只需将回调更改为parse_start_url并覆盖它即可：

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

class DownloadSpider(CrawlSpider):
    name = 'downloader'
    allowed_domains = ['bnt-chemicals.de']
    start_urls = [
        "http://www.bnt-chemicals.de",
    ]
    rules = (
        Rule(SgmlLinkExtractor(allow='prod'), callback='parse_start_url', follow=True),
    )
    fname = 0

    def parse_start_url(self, response):
        self.fname += 1
        fname = '%s.txt' % self.fname

        with open(fname, 'w') as f:
            f.write('%s, %s\n' % (response.url, response.meta.get('depth', 0)))
            f.write('%s\n' % response.body)