无法进入下一页

Question

我用 selenium 在 python 中编写了一个脚本来遍历从第一页到分页的不同页面。但是，除了一些数字外，没有下一页按钮的选项。当我单击该数字时，它会将我带到下一页。无论如何，当我尝试使用我的脚本执行此操作时，它确实会单击第二页并转到那里，但它不再滑动，我的意思是它不会继续进入第三页，而是抛出以下错误。

line 192, in check_response
    raise exception_class(message, screen, stacktrace)
selenium.common.exceptions.StaleElementReferenceException: Message: stale element reference: element is not attached to the page document

我正在尝试的脚本：

from selenium import webdriver
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()
driver.get("http://www.cptu.gov.bd/AwardNotices.aspx")
wait = WebDriverWait(driver, 10)
driver.find_element_by_id("imgbtnSearch").click()
for item in wait.until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, "#dgAwards > tbody > tr > td > a"))):
    item.click()
driver.quit()

分页编号所在的元素：

<tr align="right" valign="top" style="font-size:XX-Small;font-weight:normal;white-space:nowrap;">
        <td colspan="8"><span>Page: </span><a href="javascript:__doPostBack('dgAwards$ctl01$ctl01','')">1</a>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl02','')">2</a>&nbsp;<span>3</span>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl04','')">4</a>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl05','')">5</a>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl06','')">6</a>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl07','')">7</a>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl08','')">8</a>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl09','')">9</a>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl10','')">10</a>&nbsp;<a href="javascript:__doPostBack('dgAwards$ctl01$ctl11','')">...</a></td>
    </tr>

顺便说一句，点击主页上的“搜索”按钮会出现分页选项。

Answer 1

您无法遍历预定义元素的列表，因为在您刷新 click() 页面后，这些元素会变得陈旧

你可以试试下面的：

from selenium.common.exceptions import NoSuchElementException    

page_counter = 2
while True:
    try:
        if not page_counter % 10 == 1:
            driver.find_element_by_link_text(str(page_counter)).click()
            page_counter += 1
        else:
           driver.find_elements_by_link_text("...")[-1].click() 
           page_counter += 1
    except NoSuchElementException :
        break

这应该允许您在可能的情况下切换到下一页