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 以及 scrapy_fake_useragent

为受 cloudfare 保护的网站创建一个爬虫

我正在使用 cfscrape python extension to bypass cloudfare protection with scrapy and scrapy_fake_useragent 将随机真实 USER_AGENT 信息注入 headers。

如 cfscrape 文档所示：您必须使用相同的 user-agent 字符串来获取令牌并使用这些令牌发出请求，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 start_requests 收集的 user_agent 与 scrapy_fake_useragent 随机选择的 user_agent 不同，如您所见：

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

我在 settings.py 这个命令中定义了我的扩展：

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，那么我如何 pass/get scrapy_fake_useragent 随机给出的用户代理进入 cfscrape 扩展的 start_requests 方法？

Answer 1

在 scrapy_user_agent 开发人员的帮助下终于找到了答案。停用 settings.py 中的 'scrapy_fake_useragent.middleware.RandomUserAgentMiddleware': 400 行，然后编写此源代码：

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

在 scrapy_fake_useragent 和 cfscrape scrapy 扩展之间共享 USER_AGENT

Share USER_AGENT between scrapy_fake_useragent and cfscrape scrapy extension

python

web-scraping

scrapy

user-agent

scrapy-spider