登录Sales navigator python selenium

问题描述 投票:1回答:1

我正在努力实现一个简单的登录销售导航页面的任务,之后我可以继续尝试废弃潜在客户和帐户数据,作为一种刮取练习。在这个link登录后,我有这个脚本来完成任务

    browser = webdriver.Firefox(executable_path=r'D:\geckodriver\geckodriver.exe')

    def login_linkedin(email_id,psswd):

        sales_url = "https://www.linkedin.com/sales/login"
        browser.get(sales_url)

        elem_username = browser.find_element_by_id("username")
        #elem_username = browser.find_element_by_name("session_key")
        elem_username.send_keys(email_id)
        elem_psswd = browser.find_element_by_id("password")
        #elem_psswd = browser.find_element_by_name("session_password")
        elem_psswd.send_keys(psswd)

        elem_signin = find_element_by_class_name("login__form_action_container ").click()

    if __name__ == "__main__":

        login_linkedin("la","la")

通过检查用户名,密码和登录按钮(如下所示)字段的每个元素,我试图按名称,类名和ID查找元素,但总是以找不到元素错误结束。我能错过什么?

用户名元素

  <input id="username" name="session_key" type="text" aria-describedby="error-for-username" required="" validation="email|tel" autofocus="" aria-label="Email or Phone">

密码元素

  <input id="password" type="password" aria-describedby="error-for-password" name="session_password" required="" validation="password" aria-label="Password">

登录按钮元素

  <button class="btn__primary--large from__button--floating" type="submit" aria-label="Sign in">Sign in</button>

错误

Traceback (most recent call last):
File "D:\Freelancing\Scrap_sales_nav\code\my_code\scrap_accounts.py", line 40, in <module>
login_linkedin("la","la")
File "D:\Freelancing\Scrap_sales_nav\code\my_code\scrap_accounts.py", line 21, in login_linkedin
elem_username = browser.find_element_by_id("username")
File "C:\Python36\lib\site-packages\selenium\webdriver\remote\webdriver.py", line 360, in find_element_by_id
return self.find_element(by=By.ID, value=id_)
File "C:\Python36\lib\site-packages\selenium\webdriver\remote\webdriver.py", line 978, in find_element
'value': value})['value']
File "C:\Python36\lib\site-packages\selenium\webdriver\remote\webdriver.py", line 321, in execute
self.error_handler.check_response(response)
File "C:\Python36\lib\site-packages\selenium\webdriver\remote\errorhandler.py", line 242, in check_response
raise exception_class(message, screen, stacktrace)
selenium.common.exceptions.NoSuchElementException: Message: Unable to locate element: [id="username"]
python selenium web-scraping beautifulsoup
1个回答
2
投票

问题在于您要查找的是iFrame,因此我们需要将selenium指向iframe而不是标准网页。幸运的是,硒具有以下功能:switch_to_frame()

解决方案:

import time

browser = webdriver.Firefox(executable_path=r'D:\geckodriver\geckodriver.exe')

def login_linkedin(email_id, psswd):
        sales_url = "https://www.linkedin.com/sales/login"
        browser.get(sales_url)
        # sleeping to give the page time to load
        time.sleep(1)
        # the two lines below will put us into the iframe
        frame = browser.find_element_by_tag_name('iframe')
        browser.switch_to_frame(frame)
        # entering the username and password
        elem_username = browser.find_element_by_id("username")
        elem_username.send_keys(email_id)
        elem_psswd = browser.find_element_by_id("password")
        elem_psswd.send_keys(psswd)
        # no need for a variable name, we can just click
        browser.find_element_by_class_name("login__form_action_container ").click()

if __name__ == "__main__":
    login_linkedin('email_id', 'psswd')
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