Python Selenium 创建循环以单击页面上的链接并在每个新页面上按下按钮

Question

我对 Python 和 Selenium 还很陌生，但开始学习它。我一直在谷歌搜索如何解决这个编码问题，但找不到确切的解决方案。

我想要完成的是单击页面上的所有用户名链接，单击我被带到的页面上的关注按钮，然后返回原始页面并对其余用户名执行相同操作链接。

基本上，我想创建一个循环来执行此操作：

1. 点击第一个用户名
   • 点击关注按钮
   • 返回上一页
2. 点击第二个用户名
   • 点击关注按钮
   • 返回上一页

ETC.....通过每个链接

这是我当前的代码以及到目前为止我尝试过的代码：

from selenium import webdriver
from selenium.webdriver.common.keys import Keys

browser = webdriver.Firefox()
browser.get('thewebpage')

search = browser.find_element_by_id('getSearch')
search.click()
search.send_keys('searchitem' + Keys.RETURN)

searchitem = browser.find_elements_by_class_name("name")[0]
searchitem.click()
#I am now on the page where it shows the users

#this is where I'm getting stuck
#here's the first code I tried
links = browser.find_elements_by_link_text("#/user/")
        for link in links:
            link.click()
            follow = browser.find_element_by_class_name("followAction")
            browser.back()

#here's the second code I tried
import selenium.webdriver.support.ui as UI

def test(self):
    driver = self.driver
    wait = UI.WebDriverWait(driver, 5000)
    links = driver.find_elements_by_link_text("#/user/")
    for link in links:
        link.click()
        follow = driver.find_element_by_class_name("followAction")
        follow.click()
        driver.implicityly_wait(5)
        driver.back()

程序完成，屏幕上没有任何反应。也没有错误信息。

单击初始页面上的每个链接并单击链接将我带到的页面上的按钮，我必须进行哪些更改？

这是一个类似问题的链接。 Loop through links using Selenium Webdriver (Python)

非常感谢您的帮助。

Answer 1

已经很长时间了，但只是发布答案，如果有人仍然在某个时候检查相同类型的问题。

from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.keys import Keys
# need the below imports to work with Explicit wait
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By

browser = webdriver.Firefox()
browser.get('thewebpage')

search = browser.find_element_by_id('getSearch')
search.click()
search.send_keys('searchitem' + Keys.RETURN)

searchitem = browser.find_elements_by_class_name("name")[0]
searchitem.click()

# Here is the logic that we have to update

# Get number of users rather than the users.
userElems = len(browser.find_elements_by_link_text("#/user/"))

# iterate through each user by using the index
  # if you try to use the find_elements as shown in OP, you will get StaleElement Exception
  # because the user elements references will be refreshed when navigated to next page and
  # load back (so we have to find the elements based on index on the page every time)

for userNum in range(1,userElems):
    # this below explicit wait will make sure the script will wait max 30 sec for the next user to be clicked
    user = WebDriverWait(driver,30).until(EC.presence_of_element_located((By.XPATH,"(#/user/)[" + str(userNum) + "]")))
    # scroll user into view
    user.location_once_scrolled_into_view
    # click on user
    user.click()
    # click on follow link
    follow = WebDriverWait(driver,30).until(EC.presence_of_element_located((By.XPATH,"followAction")))
    follow.click()
    # click on browser back button
    browser.back()