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python - 试图伪造和轮换用户代理

转载 作者:太空宇宙 更新时间:2023-11-04 04:12:34 24 4
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我正在尝试伪造用户代理并在 Python 中轮换它们。
我在网上找到了一个关于如何使用 scrapy-useragents 使用 Scrapy 执行此操作的教程。包裹。
我抓取网页,https://www.whatsmyua.info/ ,为了检查我的用户代理,看看它是否与我的不同,以及它是否轮换。它是否与我的实际用户代理不同,但它不会轮换,它每次都返回相同的用户代理,我无法弄清楚出了什么问题。

设置.py

BOT_NAME = 'scrapy_javascript'

SPIDER_MODULES = ['scrapy_javascript.spiders']
NEWSPIDER_MODULE = 'scrapy_javascript.spiders'


# Crawl responsibly by identifying yourself (and your website) on the user-agent
#USER_AGENT = 'scrapy_javascript (+http://www.yourdomain.com)'

# Obey robots.txt rules
ROBOTSTXT_OBEY = True
DOWNLOADER_MIDDLEWARES = {
'scrapy_splash.SplashCookiesMiddleware': 723,
'scrapy_splash.SplashMiddleware': 725,
'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware': 810,
}

# -----------------------------------------------------------------------------
# USER AGENT
# -----------------------------------------------------------------------------

DOWNLOADER_MIDDLEWARES = {
'scrapy.downloadermiddlewares.useragent.UserAgentMiddleware': None,
'scrapy_useragents.downloadermiddlewares.useragents.UserAgentsMiddleware': 500,
}


USER_AGENTS = [
('Mozilla/5.0 (X11; Linux x86_64) '
'AppleWebKit/537.36 (KHTML, like Gecko) '
'Chrome/57.0.2987.110 '
'Safari/537.36'), # chrome
('Mozilla/5.0 (X11; Linux x86_64) '
'AppleWebKit/537.36 (KHTML, like Gecko) '
'Chrome/61.0.3163.79 '
'Safari/537.36'), # chrome
('Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:55.0) '
'Gecko/20100101 '
'Firefox/55.0'), # firefox
('Mozilla/5.0 (X11; Linux x86_64) '
'AppleWebKit/537.36 (KHTML, like Gecko) '
'Chrome/61.0.3163.91 '
'Safari/537.36'), # chrome
('Mozilla/5.0 (X11; Linux x86_64) '
'AppleWebKit/537.36 (KHTML, like Gecko) '
'Chrome/62.0.3202.89 '
'Safari/537.36'), # chrome
('Mozilla/5.0 (X11; Linux x86_64) '
'AppleWebKit/537.36 (KHTML, like Gecko) '
'Chrome/63.0.3239.108 '
'Safari/537.36'), # chrome
]

SPLASH_URL = 'http://199.89.192.74:8050'


DUPEFILTER_CLASS = 'scrapy_splash.SplashAwareDupeFilter'
HTTPCACHE_STORAGE = 'scrapy_splash.SplashAwareFSCacheStorage'

最佳答案

在这里您可以找到一个 API,它以 JSON 格式返回最常见的用户代理:
http://51.158.74.109/useragents/?format=json

我已经使用了这个工具,它会让你的用户代理列表始终更新为最新和最常用的用户代理:https://pypi.org/project/shadow-useragent/

     from shadow_useragent import ShadowUserAgent
shadow_useragent = ShadowUserAgent()

print(shadow_useragent.firefox)
# Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:67.0) Gecko/20100101 Firefox/67.0
print(shadow_useragent.chrome)
# Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/74.0.3729.169 Safari/537.36
print(shadow_useragent.safari)
# Mozilla/5.0 (Macintosh; Intel Mac OS X 10_14_5) AppleWebKit/605.1.15 (KHTML, like Gecko) Version/12.1.1 Safari/605.1.15
print(shadow_useragent.edge)
# Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/64.0.3282.140 Safari/537.36 Edge/17.17134
print(shadow_useragent.ie)
# Mozilla/5.0 (Windows NT 6.1; WOW64; Trident/7.0; rv:11.0) like Gecko
print(shadow_useragent.android)
# Mozilla/5.0 (Linux; U; Android 4.3; en-us; SM-N900T Build/JSS15J) AppleWebKit/534.30 (KHTML, like Gecko) Version/4.0 Mobile Safari/534.30
print(shadow_useragent.ipad)
# Mozilla/5.0 (iPad; CPU OS 12_3_1 like Mac OS X) AppleWebKit/605.1.15 (KHTML, like Gecko) Version/12.1.1 Mobile/15E148 Safari/604.1
print(shadow_useragent.random)
# Mozilla/5.0 (Windows NT 6.3; Win64; x64; rv:68.0) Gecko/20100101 Firefox/68.0

print(shadow_useragent.random_nomobile)
# Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/75.0.3770.90 Safari/537.36

# and the best one, random via real world browser usage statistic
print(ua.random)
# Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/75.0.3770.90 Safari/537.36

# if you want to excluse mobiles (some websites will display different pages)
print(shadow_useragent.random_nomobile)
# Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/75.0.3770.90 Safari/537.36

关于python - 试图伪造和轮换用户代理,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56082653/

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