我有一个按钮/链接,我需要点击,但无法弄明白

时间:2017-06-24 13:53:18

标签: python selenium-webdriver

我将其中一个链接伪装成我需要点击的表单上的按钮,但我无法弄清楚如何操作,这就是元素的样子:



<td id="pt1:r1:0:cf22" class="CreateQuoteBigButtonCell">
  <a id="pt1:r1:0:cl7u" class="af_commandLink" onclick="return false;" href="#">
    <table id="pt1:r1:0:tl6u" summary="" cellspacing="0" cellpadding="0" border="0">
      <tbody>
        <tr id="pt1:r1:0:rl14u">
          <td id="pt1:r1:0:cf23u" class="CreateQuoteBigButtonCell1"><img id="pt1:r1:0:i9u" class="af_image" src="/portal/images/quote/addaproduct.png"></td>
          <td id="pt1:r1:0:cf24u">
            <table id="pt1:r1:0:rl15u" summary="" cellspacing="0" cellpadding="0" border="0">
              <tbody>
                <tr>
                  <td id="pt1:r1:0:cf25u" class="CreateQuoteBigButtonCell2">Add a Product or Service</td>
                </tr>
              </tbody>
            </table>
            <table id="pt1:r1:0:rl16u" summary="" cellspacing="0" cellpadding="0" border="0">
              <tbody>
                <tr>
                  <td id="pt1:r1:0:cf26u" class="CreateQuoteBigButtonCell3">Product or service</td>
                </tr>
              </tbody>
            </table>
          </td>
        </tr>
      </tbody>
    </table>
  </a>
</td>
&#13;
&#13;
&#13;

我已经尝试过使用表ID来使用find_element_by_xpath,但它似乎无法正常工作,我还能看到其他任何选项吗?

我的代码目前看起来像这样:

browser.switch_to.frame(browser.find_element_by_css_selector("iframe[title='Quote Details']")) clickExpPack = wait.until(EC.presence_of_element_located((By.XPATH, '//*[@id="pt1:r1:0:cf16c1"]/tbody/tr/td'))) clickExpPack.click()

1 个答案:

答案 0 :(得分:0)

尝试使用以下ModuleGraph

xpath
  

driver.find_element_by_xpath("//td[@class=​'CreateQuoteBigButtonCell3‌']").click()
  

driver.find_element_by_xpath("//td[@id=​'pt1:r1:0:cf16cl‌']").click()

如果这回答你的问题,请告诉我。