Imageview on a random position

Question

I want myImageView to locate itself somewhere on a random position inside the screen, without pushing textviews away. I found this code in another thread, but I can't get it to work.

 myImageView = (ImageView) findViewById(R.id.myImageView);
        LinearLayout game = (LinearLayout)findViewById(R.id.game);

        int width = game.getWidth();
        int height = game.getHeight();

        random = new Random();

        int x = width;
        int y = height;

        int imageX = random.nextInt(x-50)+50;
        int imageY = random.nextInt(y-100)+100;

        myImageView.setX(imageX);
        myImageView.setY(imageY);

That's not the problem. The problem is that I don't know what method to use in order to place the imageview on a certain position. — Stephan Rogers, Apr 26 '16 at 17:52
I just updated the main post with my latest code that I found in another thread. According to them this code should work, but for some reason it doesn't work for me. — Stephan Rogers, May 14 '16 at 10:58

score 0 · Answer 1 · edited May 23 '17 at 11:59

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You can give margin programmatically between certain range, may be you can use device width or height for range. Check this links for setting margin dynamically on run time and get screen dimension.

edited May 23 '17 at 11:59

Community

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answered Apr 26 '16 at 17:52

Bills

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Looks helpful, but please put the relevant parts to your answer. For now this is a link only answer. – rekire Apr 26 '16 at 20:24

score 0 · Answer 2 · answered Apr 26 '16 at 17:55

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I Think first get programmatically device size then set image view width & height

WindowManager wm = (WindowManager)context.getSystemService(Context.WINDOW_SERVICE);
Display display = wm.getDefaultDisplay();

Display display = getWindowManager().getDefaultDisplay(); 
int width = display.getWidth();  // deprecated
int height = display.getHeight();  // deprecated

hear you get width & height now set image view size

android.view.ViewGroup.LayoutParams layoutParams =myImageView.getLayoutParams();
layoutParams.width = 30;
layoutParams.height = 30;
myImageView.setLayoutParams(layoutParams);

and if you still have a problem to change it's size try to put it inside of a LinearLayout and manipulate the imageView size using the layout params:

LinearLayout.LayoutParams layoutParams = new LinearLayout.LayoutParams(30, 30);
myImageView.setLayoutParams(layoutParams);

answered Apr 26 '16 at 17:55

Hardik Parmar

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If you already know that you calls are deprecated why don't you provide an alternative call? – rekire Apr 26 '16 at 20:23
I've tried your solutions but I haven't been able to solve the problem yet. I've tried using setX and setY but I can't get it to work. Here's the code: – Stephan Rogers May 09 '16 at 16:28
@StephanRogers what you get error?? – Hardik Parmar May 09 '16 at 16:29
I don't get any errors but 50% of the time the imageview position itself outside of the screen. I added the code to the main post. – Stephan Rogers May 09 '16 at 16:34
@StephanRogers get device width & height then set image view width & height – Hardik Parmar May 09 '16 at 16:43
I've done that but I can't randomize its position. Please refer to the code that I added to the main post. – Stephan Rogers May 09 '16 at 16:58
@StephanRogers Remove to your Random method to set image view x & y position use like myImageView.setMaxWidth((width / 2)); myImageView.setMaxHeight((height / 2)); – Hardik Parmar May 09 '16 at 17:10
But if I remove the Random method the imageview will have the same position all the time. – Stephan Rogers May 13 '16 at 23:29

Imageview on a random position

2 Answers2