当用户名和密码包含特殊字符时，通过Selenium进行Python HTTP基本身份验证 [英] Python HTTP Basic Authentication through Selenium when username and password contains special characters

问题描述

当尝试使用此构造使用默认的Selenium身份验证时，我有一个非常具体的问题username:[e-mail]
所以这是一个问题:如果我使用这样的密码"''asd'asd';123asd'(带有特定符号)，则traceback表示URL是不正确的.但是，如果密码仅包含数字和字母，则一切正常. 以下是一个可以正常运行的代码示例，但对于我的示例，我需要用很多特定的符号填充密码，而我无权对其进行更改.

I have a very specific problem when trying to use default Selenium auth using this construction username:[e-mail]
So here is an issue: if I use a password like this "''asd'asd';123asd' (with specific symbols) traceback will say that URL is incorrect. But if the password is just with numbers and letters -- everything is OK. Below is an example of a code that works correctly but for my example, I need to fill password with a lot of specific symbols and I don't have permission to change it.

import unittest
from selenium import webdriver


class LoggedTest(unittest.TestCase):
    @classmethod
    def setUp(inst):
        # create a new Chrome session """
        inst.driver = webdriver.Chrome()
        inst.driver.implicitly_wait(30)
        inst.driver.maximize_window()

        # navigate to the application home page """
        inst.driver.get("http://admin:[e-mail]/basic_auth")

推荐答案

如果密码包含符号，例如基本身份验证. "''asd'asd';123asd'，您需要> 翻译 字符串放入 UTF .例如:

For Basic Authentication if the Password contains symbols e.g. "''asd'asd';123asd' you need to translate the string into UTF. As an example:

username = "%21%40user" #stands for !@user
password = "%0D%0Apass" #stands for ^&pass
webpage = "something.url.com"

现在您可以使用:

url = 'http://{}:{}@{}'.format(username, password, webpage)
driver.get(url)

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