Question

所以我创建了一个使用两个操纵杆的程序。每个操纵杆都处于相对布局中，操纵杆的拇指被困在布局内，并返回我用作X和Y值的边距值。这是我的听众计划：

public boolean onTouch(View view, MotionEvent event) {
final int action = event.getAction();
final int X = (int) event.getRawX();
final int Y = (int) event.getRawY();
switch (action & MotionEvent.ACTION_MASK) {

    case MotionEvent.ACTION_DOWN: {
        RelativeLayout.LayoutParams lParams = (RelativeLayout.LayoutParams) view.getLayoutParams();
        _xDelta = X - lParams.leftMargin;
        _yDelta = Y - lParams.topMargin;
        break;
    }

    case MotionEvent.ACTION_MOVE:{
        RelativeLayout.LayoutParams layoutParams = (RelativeLayout.LayoutParams) view.getLayoutParams();
        leftM = X - _xDelta;
        //System.out.println("LeftM: " + leftM);
        topM = Y - _yDelta;
        //System.out.println("TopM: " + topM);
        layoutParams.rightMargin = -250;
        layoutParams.bottomMargin = -250;

                if(leftM < 0) {

                    leftM = 0;

                } else if (leftM > 220) {

                    leftM= 220;

                }

                if(topM < 0) {

                    topM = 0;

                } else if (topM > 220) {

                    topM = 220;

                }

                //Quadrant I
                if (topM < 110 && leftM > 110 ){

                    x = (int) (leftM - 110);
                    y = (int) (110 - topM);

                //Quadrant II
                } else if (topM < 110 && leftM < 110) {

                    x = (int) (leftM - 110);
                    y = (int) (110 - topM);

                //Quadrant III
                } else if ( topM > 110 && leftM < 110) {

                    x = (int) (leftM - 110);
                    y = (int) -(topM - 110);

                //Quadrant IV 
                } else if (topM > 110 && leftM > 110){

                    x = (int) (leftM - 110);
                    y = (int) -(topM - 110);

                //Origin
                } else {

                    layoutParams.leftMargin = leftM;
                    layoutParams.topMargin = topM;

                }

                if ((Math.pow(x, 2) + Math.pow(y, 2)) <= 12100) {

                    recentX = leftM;
                    recentY = topM;
                    layoutParams.topMargin = topM;
                    layoutParams.leftMargin = leftM;


                } else{

                    layoutParams.leftMargin = recentX;
                    layoutParams.topMargin = recentY;

                }

            switch (view.getId()) {

                case R.id.thumbL:
                    view.setLayoutParams(layoutParams);
                    joystick1.LeftY = (byte) (layoutParams.topMargin + 37);
                    joystick1.LeftX = (byte) (layoutParams.leftMargin + 37);
                    System.out.println("Left X Value: " + layoutParams.leftMargin + 37);
                    System.out.println("Left Y Value: " + layoutParams.topMargin + 37);

                case R.id.thumbR:
                    view.setLayoutParams(layoutParams);
                    joystick1.RightX = (byte) (layoutParams.leftMargin + 37);
                    System.out.println(" Right X Value: " + layoutParams.leftMargin +37);
             }
        break;

我必须将象限重新映射到正常的坐标平面（这是if语句的点）然后我有我的switch语句。我只希望每个案例在按下相应的拇指的情况下运行。然而，如果我不必握住我的右手拇指，它将打印出＃X; Right X Value＆＃34;告诉我它进入了那个案子。有什么想法吗？

Answer 1

你可以使用break。

        switch (view.getId()) {

            case R.id.thumbL:
                view.setLayoutParams(layoutParams);
                joystick1.LeftY = (byte) (layoutParams.topMargin + 37);
                joystick1.LeftX = (byte) (layoutParams.leftMargin + 37);
                System.out.println("Left X Value: " + layoutParams.leftMargin + 37);
                System.out.println("Left Y Value: " + layoutParams.topMargin + 37);
                **break;**

            case R.id.thumbR:
                view.setLayoutParams(layoutParams);
                joystick1.RightX = (byte) (layoutParams.leftMargin + 37);
                System.out.println(" Right X Value: " + layoutParams.leftMargin +37);
                **break;**
         }

Answer 2

如果您知道操纵杆占用的屏幕空间，您是否不能根据X和Y坐标检查它们？

当要求未得到满足时，程序正在进入交换机案例

2 个答案: