Selenium：处理随机弹出的窗口

We have a feature that collects customer feedback. For this , when the user logs out , a window pops up up randomly - not every time for every customer. I want to handle this in my automation code.

我们有一项收集客户反馈的功能。为此，当用户注销时，会随机弹出一个窗口 - 而不是每次都为每个客户弹出一个窗口。我想在自动化代码中处理这个问题。

Currently, at the log out, I'm expecting a window and switching to it and that code is failing when the popup window doesn't show up.

目前，在注销时，我期待一个窗口并切换到它，并且当弹出窗口没有显示时代码失败。

What's the best way to handle this .

处理这个问题的最佳方法是什么。

This is what I have so far ...

这就是我到目前为止......

    public static void waitForNumberOfWindowsToEqual(final int numberOfWindows) {


    ExpectedCondition<Boolean> expectation = new ExpectedCondition<Boolean>() {
        public Boolean apply(WebDriver driver) {
            return (driver.getWindowHandles().size() == numberOfWindows);
        }
    };

    WebDriverWait wait = new WebDriverWait(driver, BrowserFactory.explicitWait);


    wait.until(expectation);
}

3 个解决方案

#1

I would handle the absence of popup window with a try/catch. Here is an example:

我会用try / catch来处理弹出窗口的缺失。这是一个例子：

try {
    WebDriverWait winwait = new WebDriverWait(driver, 3);
    String mainWindow = driver.getWindowHandle();

    // wait for 2 windows and get the handles
    Set<String> handles = winwait.until((WebDriver drv) -> {
        Set<String> items = drv.getWindowHandles();
        return items.size() == 2 ? items : null;
    });

    // set the context on the last opened window
    handles.remove(mainWindow);
    driver.switchTo().window(handles.iterator().next());

    // close the window
    driver.close();

    // set the context back to the main window
    driver.switchTo().window(mainWindow);

} catch (TimeoutException ex) {
    System.out.println("No window present within 3 seconds");
}

#2

If possible, the ideal thing to do would be to have a look through the source to work out whether the popup window will appear, however if this isn't achievable you could take the following approach:

如果可能的话，理想的做法是查看源代码以确定是否会出现弹出窗口，但是如果无法实现这一点，您可以采取以下方法：

// Get the number of windows open before clicking the log out button.    
int numberOfWindowsBeforeLogOut = driver.getWindowHandles().size();

// Click the log out button.
logOutButton.click();

// Check how many windows are open after clicking the log out button.
int numberOfWindowsAfterLogOut = driver.getWindowHandles().size();

// Now compare the number of windows before and after clicking the log out 
// button in a condition statement.
if (numberOfWindowsBeforeLogOut < numberOfWindowsAfterLogOut) {
   // If there is a new window available, switch to it.
   driver.switchTo().window(titleOrWindowHandle);
}

#3

In case you don't get the required window, the code will throw a TimeoutException. So, put wait.until(expectation) inside a try block and catch the exception. In code,

如果您没有获得所需的窗口，代码将抛出TimeoutException。因此，将wait.until（expectation）放在try块中并捕获异常。在代码中，

try {
    wait.until(expectation);
} catch (TimeoutException ex) {
    System.out.println("Nowindow This Time");
}

#1