从随机位置获取坐标并使用它们答案

【问题标题】：get the coordinates from a random position and use them从随机位置获取坐标并使用它们
【发布时间】：2015-03-12 21:04:34
【问题描述】：

我的代码应该是这样的：

您会看到一个黑色按钮（可点击，[start_time]）和一个灰色按钮（不可点击，[buttongame1]）。 buttongame1 预测，start_time 下一步将移动到哪里。当您单击 start_time 时，start_time 将移动到 buttongame1 的 x 和 y。同时，buttongame1 将它的位置随机设置为新的。现在，当您单击 start_time 时，start_time 将移动到 buttongame1 的新 x 和 y。同时，buttongame1 随机设置它的位置，以此类推。

我已经尝试过了，但问题是，我无法从 buttongame1 获取或使用 x & y

public void onClick(View v) {

    int x0;
    int y0;
    int x1;
    int y1;
    int x2;
    int y2;
    int x3;
    int y3;
    int x4;
    int y4;
    int x5;
    int y5;
    int x6;
    int y6;
    textview1 = (TextView) findViewById(R.id.over);


    Random r = new Random();
    RelativeLayout decorView = (RelativeLayout) buttongame1.getParent();

    int screenWidth = decorView.getWidth();
    int screenHeight = decorView.getHeight();

    i++;

    if(v == gameover){
        i--;
        btncounter.setText("GAME OVER");
        start_time.setVisibility(View.GONE);
        textview1.setVisibility(View.VISIBLE);
        Animation anim = new AlphaAnimation(0.0f, 1.0f);
        anim.setDuration(50); //You can manage the time of the blink with this parameter
        anim.setStartOffset(20);
        anim.setRepeatMode(Animation.REVERSE);
        anim.setRepeatCount(Animation.INFINITE);
        textview1.startAnimation(anim);
        Toast.makeText(this, "GAME OVER", Toast.LENGTH_SHORT).show();

        finish();
    }

    if (i == 0 ) {
        btncounter.setText("0");
        buttongame1.setX(r.nextInt(screenHeight - buttongame1.getWidth()));
        buttongame1.setY(r.nextInt(screenHeight - buttongame1.getHeight()));
        int x0 = (int) buttongame1.getX();
        int y0 = (int) buttongame1.getY();
    }

    if (i == 1 ) {
        btncounter.setText("1");
        startTime = System.currentTimeMillis();
        buttongame1.setX(r.nextInt(screenHeight - buttongame1.getWidth()));
        buttongame1.setY(r.nextInt(screenHeight - buttongame1.getHeight()));
        int x1 = (int) buttongame1.getX();
        int y1 = (int) buttongame1.getY();
        start_time.setX((x1));
        start_time.setY(y1);

// ... and so on

【问题讨论】：

标签： java android button random set

【解决方案1】：

如果我理解得很好，请在更改之前从 buttongame1 中获取 X 和 Y。

     if (i == 1 ) {
            btncounter.setText("1");
             startTime = System.currentTimeMillis();
            int x1 = (int) buttongame1.getX();
            int y1 = (int) buttongame1.getY();
             buttongame1.setX(r.nextInt(screenHeight - buttongame1.getWidth()));
             buttongame1.setY(r.nextInt(screenHeight - buttongame1.getHeight()));

            start_time.setX((x1));
            start_time.setY(y1);
   }

此代码也是：

   if (i == 0 ) {
            btncounter.setText("0");
            int x0 = (int) buttongame1.getX();
            int y0 = (int) buttongame1.getY();
             buttongame1.setX(r.nextInt(screenHeight - buttongame1.getWidth()));
             buttongame1.setY(r.nextInt(screenHeight - buttongame1.getHeight()));

   }

【讨论】：