python - Scrapy CrawlSpider 不抓取第一个着陆页

Question

我是 Scrapy 的新手，我正在做一个抓取练习，我正在使用 CrawlSpider。虽然 Scrapy 框架工作得很好并且它遵循相关链接，但我似乎无法让 CrawlSpider 抓取第一个链接(主页/登陆页面)。相反，它会直接抓取规则确定的链接，但不会抓取链接所在的着陆页。我不知道如何解决这个问题，因为不建议覆盖 CrawlSpider 的解析方法。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:

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)