如何使用Python登录此网站
我正在尝试使用Python中的请求库登录到一个网站。我尝试了不同的密码,但没有成功。这是我的代码:如何使用Python登录此网站,python,login,python-requests,Python,Login,Python Requests,我正在尝试使用Python中的请求库登录到一个网站。我尝试了不同的密码,但没有成功。这是我的代码: url = 'https://www.damejidlo.cz/en/profile/prihlaseni' login_data = {'email': 'test', 'password': 'password', 'do': 'loginForm-submit'} session = requests.session() session
url = 'https://www.damejidlo.cz/en/profile/prihlaseni'
login_data = {'email': 'test',
'password': 'password',
'do': 'loginForm-submit'}
session = requests.session()
session.headers.update({'do': 'loginForm-submit'})
result = session.post(url, data=login_data)
if "test@lolco.net" in result.content:
print("example.com:", "logged in")
elif "Sign up" in result.content:
print("example.com:", "not logged in")
如何让它工作?您可以使用用户数据来测试登录。我以后会删除那个帐户。谢谢
如果我运行此代码,我会得到
example.com:notlogged in
。我想登录:D非常简单,您只需添加一个用户代理:
In [7]: import requests
In [8]: from bs4 import BeautifulSoup
In [9]: login = "https://www.damejidlo.cz/en/profile/prihlaseni"
In [10]: data = {"email": "xxxxxxxxx@gmail.com",
....: "password": "xxxxxxxx",
....: "do": "loginForm-submit"}
In [11]: h = {
....: "User-Agent": "Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/50.0.2661.75 Safari/537.36"}
In [12]: with requests.Session() as s:
....: r = s.post(login, data=data, headers=h)
....: soup = BeautifulSoup(r.content,"lxml")
....: print(soup.select_one("#snippet--user-bar a.username.user-bar__links-item").text)
....:
xxxxxxxxx@gmail.com
非常简单,您只需添加一个用户代理:
In [7]: import requests
In [8]: from bs4 import BeautifulSoup
In [9]: login = "https://www.damejidlo.cz/en/profile/prihlaseni"
In [10]: data = {"email": "xxxxxxxxx@gmail.com",
....: "password": "xxxxxxxx",
....: "do": "loginForm-submit"}
In [11]: h = {
....: "User-Agent": "Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/50.0.2661.75 Safari/537.36"}
In [12]: with requests.Session() as s:
....: r = s.post(login, data=data, headers=h)
....: soup = BeautifulSoup(r.content,"lxml")
....: print(soup.select_one("#snippet--user-bar a.username.user-bar__links-item").text)
....:
xxxxxxxxx@gmail.com
你说“不走运”是什么意思?我收到了回复200码,但我没有登录。你说“不走运”是什么意思?我收到回复200码,但我没有登录。很好,谢谢。我确实想到了这一点,但出于某种原因,我将其添加到了数据中,而不是headers param:(不用担心,数据中唯一应该包含的内容是您发布的内容,用户代理等。都是请求headersNice的一部分,谢谢。我确实想到了这一点,但出于某种原因,我将其添加到了数据中,而不是headers param:(不用担心,数据中应该包含的唯一内容是您发布的内容、用户代理等,它们都是请求头的一部分。)