Selenium / Python：找到正确位置后从长列表中选择链接

Question

该公司有 100 多个网站的列表，我正在尝试使用 Selenium webdriver 自动将用户带入该网站。我对编程相当陌生，所以如果我的问题措辞不好，请原谅我。但是，我试图从用户那里获取一个网站的名称，例如下面示例中的“Alpharetta - Cemex”，并在这么长的内容中找到它列表，然后选择该链接。通过测试，我非常确定我需要单击的元素是h3 类，该类还保存data-hmi-name下的站点名称

网站代码示例： 

我尝试过使用下面的方法，但它似乎从来没有工作过..

driver.find_element_by_css_selector("h3.tru-card-head-text uk-text-center[data-hmi-name='Alpharetta - Cemex']").click()

 

#For this one I tried to select the h3 class by searching for all those elements that has the name Alpharetta - Cemex

或者

**theCards = main.find_elements_by_tag_name("h3")** #I tried both of these declarations for theCards

**#theCards = main.find_elements_by_class_name("tru-card-wrapper")**

#then used the loop below. This obviously didn't work and it just returns an error that card.text doesn't actually exist

for card in theCards:

    #title = card.find_elements_by_tag_name("h3")

    print(card.text)

    if(card.text == theSite):

        card.click()

任何帮助或指导将不胜感激！我是 Python 编程新手，如果你能解释我做错了什么，我将永远感激不已！

Answer 1

如果您想单击单个链接（例如 Alpharetta - Cemex），您可以尝试如下：theSite = "Alpharetta - Cemex" #You can store user inputted site Name herelinkXpath = "//a[h3[contains(text(),'"+theSite +"']]"    WebDriverWait(driver, 30).until(EC.element_to_be_clickable((By.XPATH, linkXpath))).click() #This will wait for element to be clickable before it clicks如果上面不起作用。如果您的链接不在屏幕中/不可见。您可以使用java 脚本首先滚动到元素并单击，如下所示：ele = WebDriverWait(driver, 30).until(EC.presence_of_element_located((By.XPATH, linkXpath)))driver.execute_script("arguments[0].scrollIntoView();", ele )driver.execute_script("arguments[0].click();", ele )您需要导入：from selenium import webdriverfrom selenium.webdriver.common.by import Byfrom selenium.webdriver.support import expected_conditions as ECfrom selenium.webdriver.support.wait import WebDriverWait