如何在随机屏幕位置显示按钮 [英] How to display a button in random screen position

问题描述

如何在android中的随机屏幕位置显示按钮?例如，我有一个名为GO的按钮.当我单击执行"时，它将带我到第二个屏幕.第二个屏幕将在随机屏幕位置显示另一个按钮(而不是开始"按钮).我该怎么办?

How do I display a button in a random screen position in android? For example I have a button that is named GO. When I click GO, it will bring me to the second screen. That second screen will display another button (not the START button) in random screen position. How can I do that?

推荐答案

对于第二个屏幕，请使用absolute layout，但X = 0，Y = 0上的按钮

For the second screen use absolute layout, but the button on X=0, Y=0

第二个屏幕被激活后. onCreate方法

Once your second screen gets activated. onCreate Method

Button button = (Button)findViewById(R.id.my_button);
AbsoluteLayout.LayoutParams absParams = 
    (AbsoluteLayout.LayoutParams)button.getLayoutParams();

DisplayMetrics displaymetrics = new DisplayMetrics();
getWindowManager().getDefaultDisplay().getMetrics(displaymetrics);
int width = displaymetrics.widthPixels;
int height = displaymetrics.heightPixels;


Random r = new Random();

absParams.x =  r.nextInt(width ) ;
absParams.y =  r.nextInt(height );
button.setLayoutParams(absParams);

编辑:用户想知道如何编写AbsoluteLayout

EDIT User wanted to know how to write AbsoluteLayout

<AbsoluteLayout xmlns:android="http://schemas.android.com/apk/res/android"
        android:layout_width="fill_parent"
        android:layout_height="fill_parent" >
  <Button
        android:id="@+id/my_button"
        android:layout_width="100dp"
        android:layout_height="wrap_content"
        android:layout_x="0dp"
        android:layout_y="0dp"
        android:text="Yes" />
</AbsoluteLayout>

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