How to display a button in random screen position

Question

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?

get screen size, then generate random number between 0 to screen dimension then set as params to your button — Shayan Pourvatan, Mar 03 '14 at 10:53
@Shayanpourvatan How can I generate a random and set it as params? I am kind a new to android — murrmurr, Mar 03 '14 at 10:55

MDMalik · Accepted Answer · 2014-03-03T11:08:37.970

7

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

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);

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>

edited Mar 03 '14 at 11:08

answered Mar 03 '14 at 10:53

MDMalik

3,951
2
25
39

how will my xml code look like? Where will I set the button x=0, y=0? – murrmurr Mar 03 '14 at 11:09
1

@murrmurr if this solved your problem you should tick this as answer to help other people – Shayan Pourvatan Mar 03 '14 at 11:25
AbsoluteLayout is deprecated. – Mark Buikema Mar 03 '14 at 11:31
@murrmurr for setting layout params see [this](http://stackoverflow.com/questions/3144940/set-imageview-width-and-height-programmatically) – Shayan Pourvatan Mar 03 '14 at 11:44
I cant run the program yet @Shayanpourvatan it says "Force Close" I'm trying to figure out what's wrong. – murrmurr Mar 03 '14 at 13:16
I figured it out already @Shayanpourvatan. :) I forgot to code at the manifest part that's why its forced to close. But its okay now. Thank you :) – murrmurr Mar 03 '14 at 13:25

score 2 · Answer 2 · answered Feb 27 '17 at 13:54

    displaymatrics = new DisplayMetrics();
    getWindowManager().getDefaultDisplay().getMetrics(displaymatrics);

    textview.setOnTouchListener(new View.OnTouchListener(){

        Random R = new Random();
        float dx = R.nextFloat() * displaymatrics.widthPixels;
        float dy = R.nextFloat() * displaymatrics.heightPixels;

        public boolean onTouch(View view, MotionEvent event){

            if(event.getAction() == MotionEvent.ACTION_DOWN){

                        view.animate()
                                .x(dx)
                                .y(dy)
                                .setDuration(0)
                                .start();
            }
            return true;
        }
    });

AbsoluteLayout is now deprecated, You can move a View(when it's touched) in random screen position this way easily.

score 0 · Answer 3 · answered Mar 03 '14 at 10:52

If you target Icecream Sandwich and above, you can set the position of the button very easily You can use the below functions to move to a random position. You can calculate x randomly based on the screen width.

    v.setX(x);
    v.setTranslationX(translationX);

For the older phones, you can make a random animation with 0 second long.

score 0 · Answer 4 · answered Feb 23 '19 at 17:11

This is the (though Kotlin, but easy to convert to Java) code I used to animate a movement of a view on it's Y axis:

private fun moveViewRandomlyOnVerticalAxis(view: View, topMarginPixels: Int, bottomMarginPixels: Int)
{
    /* get current activity screen size in pixel **/
    val metrics = DisplayMetrics()
    windowManager.defaultDisplay.getMetrics(metrics)
    val windowHeight = metrics.heightPixels

    val viewTopPixels = view.top
    val verticalPixelsToMove = Random.nextInt(-viewTopPixels + topMarginPixels, windowHeight 
            - viewTopPixels + 1).toFloat()
    ObjectAnimator.ofFloat(view, "translationY", verticalPixelsToMove).apply {
        duration = 500
        start()
    }
}

How to display a button in random screen position

4 Answers4

Linked