问Selenium Python Click函数不工作
EN

Stack Overflow用户

提问于 2021-01-01 20:52:25

回答 2查看 99关注 0票数 0

我对这个有一个很大的问题，我现在不知道该怎么办。

所以，我正在尝试使用selenium来获取关于一种弹出窗口的信息。pop up(这就是弹出窗口，它在tiktok上)

按钮追随者的HTML元素: Followers<

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

driver = webdriver.Firefox(executable_path='./geckodriver')
driver.get("https://www.tiktok.com/@programm___r?lang=en") # on lance tiktok sur ordi 

# first thing first, just click on the button that "launch" the pop up
driver.find_element_by_class_name('jsx-1737374796.header-inbox-icon').click()
# Then, I want to click on "Followers" category but this is getting complicated here

# First idea to click on the category, check if it contains the word "followers"
if (driver.find_element_by_xpath('/html/body/div[1]/div/div[1]/div/div[3]/div[2]/div[2]/div/div[1]/div/span[5]').text) == "Followers" : # this line works, no problem 
driver.find_element_by_xpath('/html/body/div[1]/div/div[1]/div/div[3]/div[2]/div[2]/div/div[1]/div/span[5]').click() # this one doesn't work i don't know why 

# So, second idea, try with the webdriverwait
WebDriverWait(driver,20).until(EC.presence_of_element_located((By.XPATH,'/html/body/div[1]/div/div[1]/div/div[3]/div[2]/div[2]/div/div[1]/div/span[5]'))) # this works
driver.find_element_by_xpath('/html/body/div[1]/div/div[1]/div/div[3]/div[2]/div[2]/div/div[1]/div/span[5]').click() # this still doesn't work 

# Third idea, instead o xpath, css-selector 
WebDriverWait(driver,20).until(EC.presence_of_element_located((By.CSS_SELECTOR,"span.jsx-2344352055:nth-child(5)"))) # work, no problem
driver.find_element_by_css_selector("span.jsx-2344352055:nth-child(5)").click() # doesn't work neither..

# Fourth and last idea, probably the least fine, get all the elements with the class name, and only one return Followers with the .text, but still the same problem 
elements = (driver.find_elements_by_class_name("jsx-2344352055")) 
for i in range(len(elements)) : 
    if elements[i].text == "Followers" : 
         elements[i].click() # but still doesn't work

有时，它起作用了，就像我不知道为什么或如何，但有时，点击起作用了，但就像95%的时间，它不是，我真的不知道为什么

提前感谢您的帮助！

python

selenium

click

人脸识别特惠活动

精准实时的人脸检测、分析、识别等服务，在线、离线满足多样需求，可应用于智慧零售、智慧社区、在线娱乐等多种应用场景

回答 2

Stack Overflow用户

发布于 2021-01-01 21:02:15

您应该使用WebDriverWait和element_to_be_clickable expected等待按钮变为可单击状态。例如：

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

driver = webdriver.Firefox(executable_path='./geckodriver')
driver.get("https://www.tiktok.com/@programm___r?lang=en") 
WebDriverWait(driver,20).until(EC.presence_of_element_located((By.XPATH,'/html/body/div[1]/div/div[1]/div/div[3]/div[2]/div[2]/div/div[1]/div/span[5]'))) 
WebDriverWait(driver,20).until(EC.element_to_be_clickable((By.XPATH,'/html/body/div[1]/div/div[1]/div/div[3]/div[2]/div[2]/div/div[1]/div/span[5]'))).click()

票数 0

Stack Overflow用户

发布于 2021-01-01 21:15:37

请试试这个：

driver = webdriver.Firefox(executable_path='./geckodriver')
driver.get("https://www.tiktok.com/@programm___r?lang=en") # on lance tiktok sur ordi 

driver.manage().timeouts().implicitlyWait(30,TimeUnit.SECONDS) ;

# first thing first, just click on the button that "launch" the pop up
driver.find_element_by_class_name('jsx-1737374796.header-inbox-icon').click()