用于抓取单个页面的Scrapy CrawlSpider规则

Question

我基于脚本arg动态设置蜘蛛规则，并且我希望设置规则以防止对多个页面进行爬网或跟踪所有链接，并且所有爬网都取决于arg。

这是我的代码

from scrapy.exceptions import CloseSpider
from scrapy.linkextractors import LinkExtractor
from scrapy.spiders import CrawlSpider, Rule

class LinksExtractor3Spider(CrawlSpider):
    name = 'links_extractor3'
    allowed_domains = ['books.toscrape.com']
    start_urls = ['http://books.toscrape.com/']

    def __init__(self, *args, **kwargs):
        self.flow = kwargs.get('flow')
        if self.flow == 'multi':
            print('multi')
            self.rules = (
                Rule(LinkExtractor(), callback='parse_item',     follow=True),
            )
        else:
            print('single')
            print(self.start_urls[0])
            self.rules = (
                Rule(LinkExtractor(), callback='parse_item',     follow=False),
            )
        super(LinksExtractor3Spider, self).__init__(*args, **kwargs)


    def parse_item(self, response):
        item = {}
        item['title'] = response.xpath('//head/title/text()').extract()
        item['url'] = response.url
        yield item

我尝试使用Rule（LinkExtractor（），callback ='parse_item'，follow = False），但看来我不理解Rule =））Follow = False不会阻止关注。

我还尝试通过这种方式设置LinkExtractor（allow =（'start_urls [0]'））或添加另一个func parse_one_item和规则func的不同回调：

else:
    print('single')
    print(self.start_urls[0])
    self.rules = (
    Rule(LinkExtractor(), callback='parse_one_item', follow=False),
            )
    super(LinksExtractor3Spider, self).__init__(*args, **kwargs)

def parse_one_item(self, response):
    print('I COUNTER {}'.format(self.i))
    if self.i != 0:
        raise CloseSpider('finished')
    else:
        item = {}
        item['title'] = response.xpath('//head/title/text()').extract()
        item['url'] = response.url
        self.i += 1
        yield item

但是所有这些都无法正常工作。