我正在使用9图像视图我想要随机设置图像到imageview,当我点击刷新按钮,但我尝试这样它是为图像的随机分配工作,但它一次在两个(或)三个imageview中重复相同的图像.我的代码中的问题在哪
final int[] imageViews = { R.id.imgview11, R.id.imgview12, R.id.imgview13, R.id.imgview21, R.id.imgview22, R.id.imgview23, R.id.imgview31, R.id.imgview32, R.id.imgview33 }; final int[] images = { R.drawable.i1, R.drawable.i2, R.drawable.i3, R.drawable.i4, R.drawable.i5, R.drawable.i6, R.drawable.i7, R.drawable.i8, R.drawable.empty }; final ImageButton shuffle = (ImageButton) findViewById(R.id.new_puzzle); shuffle.setOnClickListener(new View.OnClickListener() { public void onClick(View view) { Random generator = new Random(); //int n = 9; //n = generator.nextInt(n); //Random random = new Random(System.currentTimeMillis()); for(int v : imageViews) { ImageView iv = (ImageView)findViewById(v); iv.setImageResource(images[generator.nextInt(images.length - 1)]); } } });
我不想重复,一个图像仅适用于一个imageview ..
使用blessenm的帖子,我写了一个你需要的类似代码.检查这是否对您有所帮助.shuffle.setOnClickListener(new View.OnClickListener() { public void onClick(View view) { Random rng = new Random(); List<Integer> generated = new ArrayList<Integer>(); for (int i = 0; i < 9; i++) { while(true) { Integer next = rng.nextInt(9) ; if (!generated.contains(next)) { generated.add(next); ImageView iv = (ImageView)findViewById(imageViews[i]); iv.setImageResource(images[next]); break; } } } } });