Python 登录 Voobly

Question

我已经阅读了数十页有关如何使用 Python 登录网页的内容，但我似乎无法使我的代码工作。我正在尝试登录一个名为“Voobly”的网站，我想知道是否有 Voobly 特有的东西使登录变得更加困难。这是我的代码:

import requests

loginURL = "https://www.voobly.com/login"
matchUrl = "https://www.voobly.com/profile/view/124993231/Matches"

s = requests.session()
loginInfo = {"username":"myUsername", "password":"myPassword"}

firstGetRequest = s.get(loginURL) # Get the login page using our session so we save the cookies

postRequest = s.post(loginURL,data=loginInfo) # Post data to the login page, the data being my login information

getRequest = s.get(matchUrl) # Get content from a login - restricted page

response = getRequest.content.decode() # Get the actual html text from restricted page

if "Page Access Failed" in response: # True if I'm blocked
    print("Failed")
else: # If I'm not blocked, I have the result I want
    print("Worked!") # I can't achieve this

Answer 1

如上所述in the comments ，登录表单提交到/login/auth。但是，cookie 是从 /login URL 生成的。

使用以下代码:

form = {'username': USERNAME, 'password': PASSWORD}

with requests.Session() as s:
    # Get the cookie
    s.get('https://www.voobly.com/login')
    # Post the login form data
    s.post('https://www.voobly.com/login/auth', data=form)
    # Go to home page
    r = s.get('https://www.voobly.com/welcome')
    # Check if username is in response.text
    print(USERNAME in r.text)
    # True

    r2 = s.get('https://www.voobly.com/profile/view/124993231/Matches')
    if "Page Access Failed" in r2.text:
        print("Failed")
    else:
        print("Worked!")
    # Worked!

注意:登录根本不需要转到主页部分。仅用于表明登录成功。