在scrapy_fake_useragent和cfscrape scrapy扩展之间共享USER_AGENT

Question

我正在尝试使用cfscrape ， privoxy和tor和scrapy_fake_useragent为受云端保护的网站创建一个刮板

我正在使用cfscrape python扩展名 ，以scrapy和scrapy_fake_useragent绕过cloudfare保护，以将随机的真实USER_AGENT信息注入标头中。

如cfscrape文档所述： 必须使用相同的用户代理字符串来获取令牌以及使用这些令牌进行请求，否则Cloudflare会将您标记为机器人。

To collect cookie needed by `cfscrape`, i need to redefine the `start_request` function into my spider class, like this : 

    def start_requests(self):
        cf_requests = []
        for url in self.start_urls:
            token, agent = cfscrape.get_tokens(url)
            self.logger.info("agent = %s", agent)
            cf_requests.append(scrapy.Request(url=url,
                                              cookies= token,
                                              headers={'User-Agent': agent}))
        return cf_requests

我的问题是， user_agent通过收集start_requests是不一样的user_agent随机选定scrapy_fake_useragent ，你可以看到：

017-01-11 12:15:08 [airports] INFO: agent = Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:41.0) Gecko/20100101 Firefox/41.0
2017-01-11 12:15:08 [scrapy.core.engine] INFO: Spider opened
2017-01-11 12:15:08 [scrapy.extensions.logstats] INFO: Crawled 0 pages (at 0 pages/min), scraped 0 items (at 0 items/min)
2017-01-11 12:15:08 [scrapy.extensions.telnet] DEBUG: Telnet console listening on 127.0.0.1:6023
2017-01-11 12:15:08 [scrapy_fake_useragent.middleware] DEBUG: Assign User-Agent Mozilla/5.0 (Macintosh; Intel Mac OS X 10_7_3) AppleWebKit/534.55.3 (KHTML, like Gecko) Version/5.1.3 Safari/534.53.10 to Proxy http://127.0.0.1:8118

我按以下顺序在settings.py定义了扩展名：

RANDOM_UA_PER_PROXY = True
HTTPS_PROXY = 'http://127.0.0.1:8118'
COOKIES_ENABLED = True

DOWNLOADER_MIDDLEWARES = {
    'scrapy.downloadermiddlewares.useragent.UserAgentMiddleware': None,
    'scrapy_fake_useragent.middleware.RandomUserAgentMiddleware': 400,
    'flight_project.middlewares.ProxyMiddleware': 100,
    'scrapy.downloadermiddlewares.httpproxy.HttpProxyMiddleware':110,
    }

我需要相同的user_agent ，那么如何将scrapy_fake_useragent随机给定的用户代理传递/获取到cfscrape扩展的start_requests方法中？

Answer 1

scrapy_user_agent在scrapy_user_agent开发人员的帮助下找到了答案。 在settings.py 'scrapy_fake_useragent.middleware.RandomUserAgentMiddleware': 400 ，然后编写以下源代码：

class AirportsSpider(scrapy.Spider):
    name = "airports"
    start_urls = ['https://www.flightradar24.com/data/airports']
    allowed_domains = ['flightradar24.com']

    ua = UserAgent()
    ...

    def start_requests(self):
        cf_requests = []
        user_agent = self.ua.random
        self.logger.info("RANDOM user_agent = %s", user_agent)
        for url in self.start_urls:
            token , agent = cfscrape.get_tokens(url,user_agent)
            self.logger.info("token = %s", token)
            self.logger.info("agent = %s", agent)

            cf_requests.append(scrapy.Request(url=url,
                                          cookies= token,
                                          headers={'User-Agent': agent}))
        return cf_requests