Selenium Timeoutexception错误

Question

仍在处理这个Instagram问题，我真的需要你的帮助。

这是我的代码：

input_button = wait(browser,10).until(EC.element_to_be_clickable((By.XPATH, 
'//button[@class ="chBAG"]')))

action=ActionChains(browser)
action.move_to_element(input_button)
action.click()
action.perform()

这是HTML：

<button class="chBAG">Fermer</button>

但我得到了：

selenium.common.exceptions.TimeoutException: Message: 

有人可以帮我解决这个问题吗？

THX

Answer 1

由于输入字段的泛型类，您收到此错误。每次打开页面时，都会随机生成类名。那么如何解决呢？例如：

想象一下你想要登录，你必须：

1. 点击email输入字段
2. 类型信息
3. 点击password输入字段
4. 类型信息
5. 点击Log in按钮

示例代码可能是这样的：

# xpath will work every time because it is static
email_input = wait(browser,10).until(EC.element_to_be_clickable((By.XPATH, 
'//form/div[1]/div/div/input'))) # locate email input
email_input =ActionChains(browser)
email_input.move_to_element(email_input)
email_input.click()
email_input.sendKeys("email")
email_input.perform()

password_input = wait(browser,10).until(EC.element_to_be_clickable((By.XPATH, 
'//form/div[2]/div/div/input'))) # locate password input
password_input =ActionChains(browser)
password_input.move_to_element(password_input)
password_input.click()
email_input.sendKeys("password")
password_input.perform()

login_button = wait(browser,10).until(EC.element_to_be_clickable((By.XPATH, 
'//span/button'))) # locate login button
login_button_action=ActionChains(browser)
login_button_action.move_to_element(login_button )
login_button_action.click()
login_button_action.perform()

要在搜索栏中搜索内容，您必须执行以下操作：

1. 点击search输入字段
2. 类型信息
3. 等到results加载
4. 点击results
中的一个

代码：

import time # will be need below

search_input = wait(browser,10).until(EC.element_to_be_clickable((By.XPATH, 
"//input[@placeholder = 'Search']"))) # locate search input
search_input =ActionChains(browser)
search_input.move_to_element(search_input)
search_input.click()
search_input.sendKeys("fermer")
search_input.perform()

time.sleep(5) # wait 5 seconds until dropdown will appear

dropdown_menu = wait(browser,10).until(EC.element_to_be_clickable((By.XPATH, 
"//a[@href = '/explore/tags/fermermaid/']"))) # locate fermeraid
dropdown_menu = ActionChains(browser)
dropdown_menu.move_to_element(dropdown_menu)
dropdown_menu.click()
dropdown_menu.perform()

Answer 2

根据您的要求，您可以使用以下代码：

它会搜索＆＃34; 这是测试＆＃34;在Instagram的搜索栏中的字符串。

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

driver = webdriver.Chrome(executable_path = r'D:/Automation/chromedriver.exe')
driver.get("https://www.instagram.com/accounts/login/")

username = WebDriverWait(driver, 10).until(
EC.element_to_be_clickable((By.NAME, "username")))

password = WebDriverWait(driver, 10).until(
EC.element_to_be_clickable((By.NAME, "password")))

username.send_keys("your username")
password.send_keys("your password")
driver.find_element_by_tag_name('button').click()

search = WebDriverWait(driver, 10).until(
EC.element_to_be_clickable((By.XPATH, "//span[text()='Search']")))

search.click()
driver.find_element_by_xpath("//div[@role='dialog']/preceding-sibling::input").send_keys("This is testing")