Java/Selenium WebDriver : Focus don't skip to ELSE block if a webelement is not found on page

Question

我正在使用 Java 开发 Selenium WebDriver，我需要检查输入无效用户名和密码时的场景，应用程序应该抛出警告消息。如果输入正确的凭据，则不会显示警告消息。我在下面编写了一段代码来验证有关错误登录凭据的警告消息。但是，当我输入有效的详细信息时，我的代码不会跳到 else block ，而是失败说它无法识别元素 ("//*[@class=' Small-9 Small-pull-1 列内容']").

代码:

try {           
             if (new WebDriverWait(driver, 10).until(ExpectedConditions.textToBePresentInElementLocated(By.xpath("//*[@class='small-9 small-pull-1 column content']"), "No Password Found for"))){

            String text = driver.findElement(By.xpath("//*[@class='small-9 small-pull-1 column content']")).getAttribute("innerHTML");
            System.out.println(text);

            if(text.contentEquals("No Password Found for")){
                    driver.navigate().refresh();

                Assert.fail("Unable to login to application");
            }
            }

             else if(new WebDriverWait(driver, 10).until(ExpectedConditions.textToBePresentInElementLocated(By.xpath("//*[@class='small-9 small-pull-1 column content']"), "Your Online/Mobile Banking User ID has been blocked. Please go to “Forgot Password” option to unblock it."))){

               String retext = driver.findElement(By.xpath("//*[@class='small-9 small-pull-1 column content']")).getAttribute("innerHTML");

            System.out.println(retext);

            if(retext.contentEquals("Your Online/Mobile Banking User ID has been blocked. Please go to “Forgot Password” option to unblock it.")) {
                driver.navigate().refresh();

                Assert.fail("Unable to login to application");
            }


        }       

             else if (new WebDriverWait(driver, 10).until(ExpectedConditions.textToBePresentInElementLocated(By.xpath("//*[@class='small-9 small-pull-1 column content']"), "Time Out"))){
            String timeouttext = driver.findElement(By.xpath("//*[@class='small-9 small-pull-1 column content']")).getAttribute("innerHTML");

            if(timeouttext.contentEquals("You have specified an invalid User Name or Password. Please check and try again")){
                driver.navigate().refresh();

                                    Assert.fail("Unable to login to application");      }


        }

带有失败消息的 HTML block :

<div class="column">
    <div class="form-field ">
        <div class="popup-login-error visible -full error" id="j_idt65">
            <div class="row"><span class="icon-cancel-thin color-white" messagepopupclose=""></span>
                <div class="small-2 column icon"><i class="icon-exclamation-circle"></i></div>
                <div class="small-9 small-pull-1 column content">You have specified an invalid User Name or Password. Please check and try again</div>
                <div class="small-9 small-pull-1 column content">No Password Found for </div>
            </div>
        </div>
    </div>
</div>

Answer 1

我已经优化了您的代码块，如下所示:

• 在任何情况下都必须将 driver.findElement(By.name("subscribed")).click(); 从 try{} block 中取出 click()它。
• 引发WebDriverWait，使文本无效的用户名或密码出现在HTML DOM
• 保持错误文本用户名或密码无效的xpath不变。
• 将 getText() 替换为 getAttribute("innerHTML")
• 保持 contentEquals() 和 Assert.fail() 逻辑不变。
• 删除了 driver.navigate().refresh();，因为它可能会引发 StaleElementReferenceException
• 如果您的尝试{}失败错误消息，您的程序将打印显示仪表板

• 这是您的工作代码块:

  //code block
  driver.findElement(By.name("submitted")).click();
  try {
      new WebDriverWait(driver, 10).until(ExpectedConditions.textToBePresentInElementLocated(By.xpath("//*[@class='small-9 small-pull-1 column content']"), "invalid User Name or Password"));
      String text = driver.findElement(By.xpath("//*[@class='small-9 small-pull-1 column content']")).getAttribute("innerHTML");
      System.out.println(text);
      if (text.contentEquals("You have specified an invalid User Name or Password. Please check and try again") || text.contentEquals("Time Out") || text.contentEquals("Your Online/Mobile Banking User ID has been blocked. Please go to “Forgot Password” option to unblock it.")) 
      {
          Assert.fail("Unable to login to application");
      }
  } catch (Exception e) {
      System.out.println("DAashboard is displayed");
      // rest of code block
  }