Appium MobileElement滑动返回未知服务器错误

Question

我试图刷MobileElement，但它给服务器端错误。这里的原因可能是什么？

代码：

 MobileElement mb= (MobileElement)driver1.findElement(By.xpath("//android.widget.ListView[@index='0'][@resource-id='android:id/list']"));

 mb.swipe(SwipeElementDirection.LEFT, 1000);

错误：

org.openqa.selenium.WebDriverExcture：在处理命令时发生了未知的服务器端错误。（警告：服务器未提供任何stackTrace信息）命令持续时间或超时：16毫秒

Answer 1

使用AppiumDriver更好地在基类中实现这一点

public void swipe(int startx, int starty, int endx, int endy, int duration)

称之为：

driver.swipe((int)screenWidth(), (int)screenHeight()*0.6), (int)screenWidth(), (int)screenHeight()*0.4), 0);

Answer 2

您可以尝试使用这段代码为所有移动设备进行动态滑动：

Dimension dimension = driver.manage().window().getSize();
int width = dimension.getWidth();
int height = dimension.getHeight();
switch(direction) {
    case "right":
        driver.swipe((int) (width*(0.20)), (int) (height*(0.50)), (int)(width*(0.80)), (int) (height*(0.50)), 6000);
    break;
    case "left":
        driver.swipe((int) (width*(0.80)), (int) (height*(0.50)), (int) (width*(0.20)), (int) (height*(0.50)), 6000);
    break;
    case "up":
        driver.swipe((int) (width*(0.50)), (int) (height*(0.70)), (int) (width*(0.50)), (int) (height*(0.30)), 6000);
    break;
    default:
        driver.swipe((int) (width*(0.50)), (int) (height*(0.30)), (int) (width*(0.50)), (int) (height*(0.70)), 6000);
    break;
}

Answer 3

试试这个，这将工作：

TouchAction action = new TouchAction(driver);

int startY2 = element1.getLocation().getY() + (element.getSize().getHeight() / 2);

int startX2 = element1.getLocation().getX() + (element.getSize().getWidth() / 2);

int endX2 = element2.getLocation().getX() + (element2.getSize().getWidth() / 2);

int endY2 = element2.getLocation().getY() + (element2.getSize().getHeight() / 2) - (element2.getSize().getHeight()/2);


action.press(startX2, startY2).waitAction(2000).moveTo(endX2, endY2).release().perform();