硒单击GDPR单击按钮

问题描述 投票:1回答:1

我正在尝试以编程方式单击在ubuntu服务器上运行脚本的按钮以访问网站内容。我正在尝试通过recipies

到达该站点

这是我的代码:


driver = webdriver.Firefox(executable_path="/home/ubuntu/.linuxbrew/Cellar/geckodriver/0.26.0/bin/geckodriver",options=options)
data = []


# In[34]:

# click GDPR full-width banner 
start_time = datetime.now()
driver.get("http://allrecipes.co.uk/")
time.sleep(10)

# gdpr_button = driver.find_element_by_link_text("Continue")
#gdpr_button = driver.find_element_by_xpath('//button[text()="Continue"]')

driver.find_element_by_xpath("//input[@style='order:2' and @onclick='sendAndRedirect()']").click()

# //input[@onclick='sendAndRedirect()']
# <button style="order:2" onclick="sendAndRedirect();">Continue</button>

但是,我无法到达元素并得到

selenium.common.exceptions.TimeoutException: Message: connection refused

如何访问GDPR表格中的“继续”按钮?感谢您的帮助

javascript python selenium web-scraping geckodriver
1个回答
0
投票

所需元素是动态元素,因此要在元素上定位/ click(),需要为element_to_be_clickable()引入WebDriverWait,并且可以使用以下任何一个Locator Strategies

  • 使用CSS_SELECTOR

    WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "button[onclick^='sendAndRedirect']"))).click()
    
  • 使用XPATHinnerText事件:

    WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//button[text()='Continue']"))).click()
    
  • 使用XPATHonclick事件:

    WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//button[starts-with(@onclick, 'sendAndRedirect')]"))).click()
    
  • Note:您必须添加以下导入:

    from selenium.webdriver.support.ui import WebDriverWait
    from selenium.webdriver.common.by import By
    from selenium.webdriver.support import expected_conditions as EC
    
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