How to get the Touch position in android?
@Override
public boolean onTouchEvent(MotionEvent event)
{
int x = (int)event.getX();
int y = (int)event.getY();
switch (event.getAction()) {
case MotionEvent.ACTION_DOWN:
case MotionEvent.ACTION_MOVE:
case MotionEvent.ACTION_UP:
}
return false;
}
The three cases are so that you can react to different types of events, in this example tapping or dragging or lifting the finger again.
Supplemental answer
Given an OnTouchListener
private View.OnTouchListener handleTouch = new View.OnTouchListener() {
@Override
public boolean onTouch(View v, MotionEvent event) {
int x = (int) event.getX();
int y = (int) event.getY();
switch (event.getAction()) {
case MotionEvent.ACTION_DOWN:
Log.i("TAG", "touched down");
break;
case MotionEvent.ACTION_MOVE:
Log.i("TAG", "moving: (" + x + ", " + y + ")");
break;
case MotionEvent.ACTION_UP:
Log.i("TAG", "touched up");
break;
}
return true;
}
};
set on some view:
myView.setOnTouchListener(handleTouch);
This gives you the touch event coordinates relative to the view that has the touch listener assigned to it. The top left corner of the view is (0, 0)
. If you move your finger above the view, then y
will be negative. If you move your finger left of the view, then x
will be negative.
int x = (int)event.getX();
int y = (int)event.getY();
If you want the coordinates relative to the top left corner of the device screen, then use the raw values.
int x = (int)event.getRawX();
int y = (int)event.getRawY();
Related
- How are Android touch events delivered?
@Override
public boolean onTouch(View v, MotionEvent event) {
float x = event.getX();
float y = event.getY();
return true;
}