如何使用WebDriver处理Appium中的弹出窗口

Question

I unable to click Open button and getting an exception as "No such element exception". I tried all XPath and id but same exception I am getting. I have tried getContextHandles() and it returns size as one. it means control cannot switch. Kindly help me to click the "Open" button.

Below are the xPath i tried.

By.name("Open");
By.xpath("//android.widget.Button[@index='1']");
By.xpath("//android.widget.Button[@resource-id='org.hola:id/btn_open']");
By.xpath("//android.widget.Button[@resource-id='org.hola:id/btn_open' and @index='1' and @text='Open']");

Answer 1

如果启动任何应用程序后都会弹出此窗口，则必须在代码中设置“ appActivity”和“ appPackage”功能。

Answer 2

Try MobileBy instead of By

MobileBy.name("Open");
MobileBy.xpath("//android.widget.Button[@index='1']");
MobileBy.xpath("//android.widget.Button[@resource-id='org.hola:id/btn_open']");
MobileBy.xpath("//android.widget.Button[@resource-id='org.hola:id/btn_open' and @index='1' and @text='Open']");

Answer 3

Locator can be identified by using UIAutomator viewer tool. Following code setup use to handles the pop-up window in Appium

WebDriver driver;
@BeforeTest
static void setup() throws MalformedURLException{
    DesiredCapabilities cap = new DesiredCapabilities();
    cap.setCapability(CapabilityType.PLATFORM, "Android");
    cap.setCapability(CapabilityType.VERSION, "Android-version"); 
    cap.setCapability("deviceName", "connected device name");
    cap.setCapability("appPackage", "App package name");
    cap.setCapability("appActivity", "App activity");
    driver = new RemoteWebDriver(new URL("http://127.0.0.1:4723/wd/hub"), cap); 
}
@Test
static void Selendroid() throws InterruptedException{
    Thread.sleep(2000);
    driver.findElement(By.name("Locator")).click();
    System.out.println(driver.findElementById("android:id/message").getText());
    driver.findElement(By.name("Locator")).click();
}

For Browser alerts use driver.switchTo().alert() method to Handle