Question

I’m having trouble in making unrepeating random images on my Android App.

Let’s say on my App I have this TOP 3 Sections, and I have 13 pictures. I need a method to generate 3 random images from the 13 pics on @drawable I have, but the 3 pictures should be different from one another.

For example: it can’t be img1,img1,img2 or img1,img1,img1 on the TOP 3 section but it should be img1,img2,img3.

So far, I have this method in hand:

final Button imgView = (Button)findViewById(R.id.top1);
        Random rand = new Random();            
        int rndInt = rand.nextInt(13) + 1; 
        String imgName = "img" + rndInt;
        int id = getResources().getIdentifier(imgName, "drawable", getPackageName());  
        imgView.setBackgroundResource(id);

On my layout, I specified one of the image ID on the TOP 3 section as top1. I still have the id top2, and top3 for the TOP 3 section. The above code will look up to my drawable images, which have the names ‘img1.jpg’, ‘img2.jpg’, ‘img3.jpg’ , ‘img4.jpg’, etc.

I’m open to any kind of solution. If somebody could improvise my code so 3 pictures won’t be the same with each other that would be perfect. But if any of you can offer a better solution with a new method, I’m open for it.

Thank you.

Answer 1

I’d suggest you to store your image ids in List, use shuffle
and retrieve the first element by removing.

//the initial list of images
final List<String> images = new ArrayList<String>();
for (int i=1; i<=13; i++){
images.add("img"+i);
}
//the method that retrieves random value
Collections.shuffle(images, new Random());
String randomImage = images.remove(0);

EDIT:

//this code should be inoked on every image add
final Button imgView = (Button)findViewById(R.id.top1);
Collections.shuffle(images, new Random());
String imgName = images.remove(0);
int id = getResources().getIdentifier(imgName, "drawable", getPackageName());  
imgView.setBackgroundResource(id);

The initialization of list should be performed once, outside of image add method (somewhere in onCreate probably.