从抽象类访问Android Accessibility.Gesture

Question

我正在尝试在屏幕的某一X Y部分执行轻触操作。我遇到的问题是，当我直接从我的抽象类调用它时，dispatchGesture不会触发。如何从抽象类中直接调用它？

到目前为止，我知道如果我在抽象类中设置一个静态变量，我就可以访问它，但这意味着必须触发新的事件，而不是使用触发的过去的事件，这有点慢或过时：

public class AutoService extends AccessibilityService {
public static boolean canDoTask = false;
public static int SX = 100;
public static int SY = 100;

@Override
public void onAccessibilityEvent(AccessibilityEvent event) {
 AccessibilityNodeInfo source = event.getSource();
    String packageName = Tools.getPackage(source);

  if (canDoTask){
      pressLocation(SX, SY);
   }
}

public void pressLocation(int x, int y){
    Log.e(TAG, "PRESS LOCATION : " + x +":"+ y);
    GestureDescription.Builder builder = new GestureDescription.Builder();
    Path p = new Path();
    p.moveTo(x, y);
    builder.addStroke(new GestureDescription.StrokeDescription(p, 150L, 50L));
    GestureDescription gesture = builder.build();
    dispatchGesture(gesture, new GestureResultCallback() {
        @Override
        public void onCompleted(GestureDescription gestureDescription) {
            super.onCompleted(gestureDescription);
            Log.e(TAG, "DISPACHED");
        }

        @Override
        public void onCancelled(GestureDescription gestureDescription) {
            super.onCancelled(gestureDescription);
            Log.e(TAG, "CANCELLED");
        }
    }, null);

    Log.e(TAG, "END OF PRESS");
    canDoTask = false;
}
}

这是我想要从其中调用方法的第二个类：

public abstract class MyOtherClass
{
    // SOME CODE HERE TO CHECK VARIABLES



    // THIS IS WHERE I WANT TO CALL pressLocation at
    public static void processTask()
    {
         canDoTask = true;
    }
}

我在一个论坛上看到有人使用onServiceConnect()实现了这一点，但他们没有任何示例代码来说明如何做到这一点，或者确切地说你应该采用哪些方法。

protected void onServiceConnected()
{
    super.onServiceConnected();
}

Answer 1

您可以创建一个AccessibilityService实例并通过它调用方法。

1.创建AccessibilityService的静态字段

public class AutoService extends AccessibilityService {
    private static AutoService mAutoService; 
}

2.在onServiceConnected()中初始化

public class AutoService extends AccessibilityService {
    @Override
    protected void onServiceConnected() {
        super.onServiceConnected();
        mAutoService = this;
    }

...
}

3.创建获取实例的方法：

public class AutoService extends AccessibilityService {
    public static MyAccessibilityService getInstance(){
        return mAutoService;
    }

...

}

4.现在在您的抽象类中：

public abstract class MyOtherClass {
    AutoService autoService = AutoService.getInstance();
    if(autoService != null) {
        autoService.pressLocation(x, y);
    }
}