如何在python中处理超时异常

Question

我正在编写一个机器人，该机器人将在www.quora.com上关注用户。 以下是我在超时异常中使用的代码部分：

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
import time
import urllib

driver = webdriver.Firefox()
driver.get('http://www.quora.com/')
time.sleep(10)

wait = WebDriverWait(driver, 10)

form = driver.find_element_by_class_name('regular_login')
time.sleep(10)
#add explicit wait

username = form.find_element_by_name('email')
time.sleep(10)
#add explicit wait

username.send_keys('abc@gmail.com')
time.sleep(30)
#add explicit wait

password = form.find_element_by_name('password')
time.sleep(30)
#add explicit wait

password.send_keys('def')
time.sleep(30)
#add explicit wait

password.send_keys(Keys.RETURN)
time.sleep(30)

#search = driver.find_element_by_name('search_input')
search = wait.until(EC.presence_of_element_located((By.XPATH, "//form[@name='search_form']//input[@name='search_input']")))

search.clear()
search.send_keys('Kevin Rose')
search.send_keys(Keys.RETURN)

link = wait.until(EC.presence_of_element_located((By.LINK_TEXT, "Kevin Rose")))
link.click()
#Wait till the element is loaded (Asynchronusly loaded webpage)

handle = driver.window_handles
driver.switch_to.window(handle[1])
#switch to new window 

element = WebDriverWait(driver, 2).until(EC.presence_of_element_located((By.PARTIAL_LINK_TEXT, "Followers")))
time.sleep(30)
element.click()
#goes to Kevin Rose followers page
time.sleep(30)

button = driver.find_elements_by_xpath("//a[contains(text(), 'Follow')]")
#Locate follow button on the page
no_of_followers = len(button)
#total number of unfollowed users
print no_of_followers


    while(no_of_followers > 0):
    # execute only if there are unfollowed users on page

        count = 1

        while(count < no_of_followers):

            time.sleep(30)
            link = wait.until(EC.presence_of_element_located((By.LINK_TEXT, "Follow")))
            time.sleep(30)
            link.click()
            time.sleep(30)
            print count
            count = count + 1


        time.sleep(30)
        driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
        time.sleep(30)
        button = driver.find_elements_by_xpath("//a[contains(text(), 'Follow')]")
        time.sleep(30)
        no_of_followers = len(button)

执行代码后，成功执行一次后，我在内部循环中收到“ TimeoutException”错误。

我该如何解决？

追溯：

追溯（最近一次通话）：文件“ C：\\ Python27 \\ quorabot7”，第72行，链接= wait.until（EC.presence_of_element_located（（By.LINK_TEXT，“ Follow”））））文件“ C：\\ Python27 \\ lib \\ site-packages \\ selenium \\ webdriver \\ support \\ wait.py“，第71行，直到引发TimeoutException（message）TimeoutException：消息：”

Answer 1

您收到一个TimeoutException，因为Selenium在您设置为等待的时间内找不到该元素。 这意味着您的定位器策略不正确。

我还没有测试过其他定位器，但是如果确实是内部循环失败了，我的解决方案将在下面。

在凯文·哈特（Kevin Hart）的页面上浏览了DOM之后，我可以看到您感兴趣的按钮是：

<a class="follow_button with_count" href="#" action_click="UserFollow" id="__w2_Mab4s9V_follow_user">Follow<span class="count">43.8k</span></a>

您应该尝试这样：

link = wait.until(EC.presence_of_element_located(\
       (By.className, "follow_button with_count")))

或这个：

link = wait.until(EC.presence_of_element_located(\
       (By.XPATH, '//a[@action_click="UserFollow"]')))

Answer 2

正确的是By.CLASS_NAME ，而不是By.ClassName