我曾经有过类似的情况。所以我把我所有的屏幕元素都放在了 PageObject 设计模式中。因此,在每次滑动之后,我都会重新初始化屏幕,并通过该逻辑重新初始化其所有元素,并且可以肯定的是,我已经检查了某个元素是否已加载到该屏幕上,因此在伪代码中将如下所示:
public void checkSwipeScreen(){
swipe(driver, DIRECTION.LEFT, 5);
Screen screen = new Screen(driver);
Assert.assertTrue(screen.elementIsLoaded(button), "Screen is not loaded properly");
}
因此,如果在您的新屏幕上加载了所有/一个元素,则滑动是成功的。
使用最后一个参数您可以控制滑动是否更长(持续时间长)。
如果你喜欢,这里是滑动方法:
public static void swipe(MobileDriver driver, DIRECTION direction, long duration) {
Dimension size = driver.manage().window().getSize();
int startX = 0;
int endX = 0;
int startY = 0;
int endY = 0;
switch (direction) {
case RIGHT:
startY = (int) (size.height / 2);
startX = (int) (size.width * 0.90);
endX = (int) (size.width * 0.05);
new TouchAction(driver)
.press(startX, startY)
.waitAction(Duration.ofMillis(duration))
.moveTo(endX, startY)
.release()
.perform();
break;
case LEFT:
startY = (int) (size.height / 2);
startX = (int) (size.width * 0.05);
endX = (int) (size.width * 0.90);
new TouchAction(driver)
.press(startX, startY)
.waitAction(Duration.ofMillis(duration))
.moveTo(endX, startY)
.release()
.perform();
break;
case UP:
endY = (int) (size.height * 0.70);
startY = (int) (size.height * 0.30);
startX = (size.width / 2);
new TouchAction(driver)
.press(startX, startY)
.waitAction(Duration.ofMillis(duration))
.moveTo(endX, startY)
.release()
.perform();
break;
case DOWN:
startY = (int) (size.height * 0.70);
endY = (int) (size.height * 0.30);
startX = (size.width / 2);
new TouchAction(driver)
.press(startX, startY)
.waitAction(Duration.ofMillis(duration))
.moveTo(startX, endY)
.release()
.perform();
break;
}
}
和枚举
public enum DIRECTION {
DOWN, UP, LEFT, RIGHT;
}
希望这会有所帮助,