Web 应用程序中的 Appium:无法在通知弹出窗口中点击允许权限按钮

问题描述

当我打开网络应用程序时，我收到一个弹出窗口.我试图通过两种方式点击允许"按钮:

When I am opening the web app, I am getting a pop up. I'm trying to click on 'ALLOW' button in two ways:

1) 当我添加权限时:

1) When I add permissions:

caps.setCapability("autoGrantPermissions", true);
caps.setCapability("autoAcceptAlerts", true);
......
driver.switchTo().alert().accept();

什么都没发生((

2) 当我尝试通过 XPath 找到它时:

2) When I try to find it by XPath:

driver.findElement(By.xpath("//android.widget.Button[@text='Allow']")).click();

我收到一个错误:

Exception in thread "main" org.openqa.selenium.NoSuchElementException: no such element: Unable to locate element: {"method":"xpath","selector":"//android.widget.Button[@text='Allow']"}

这是我在 UI Automator Viewer 中的截图:

This is my screenshot from UI Automator Viewer :

我找到了这个帖子:在 Appium 中点击权限提醒的允许按钮后无法点击链接?但这对我没有帮助.

I found this post:Unable to tap the link after tap on Allow button of permission alert in Appium?but it didn't help me.

推荐答案

第一:功能您尝试使用的是 仅限IOS

First: capabilities you were trying to use are for IOS only

在 Android 上，您必须通过 findElement 找到弹出窗口并自己关闭它们.

On Android you have to find popup via findElement and close them yourself.

第二:由于您为网络应用程序启动 Appium 会话，因此在搜索原生弹出窗口之前，您必须切换上下文:

Second: since you start Appium session for web application, before searching for native popups you must switch the context:

    String webContext = driver.getContext();
    Set<String> contexts = driver.getContextHandles();
    for (String context: contexts){
        if (context.contains("NATIVE_APP")){
            driver.context(context);
            break;
        }
    }
    driver.findElement(By.id("android:id/button1")).click();

不要忘记将上下文切换回网络以继续:

Don't forget to switch context back to web to continue:

    driver.context(webContext);

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