java - 在某些条件下撤消微调器选择

Question

也许是一个奇怪的问题。我有两个微调器，它们都适应同一个列表中的值。最初，我将它们设置为不同的位置。我需要它们始终不同，即用户永远无法在第一个 Spinner 中选择已在第二个中选择的 String ，反之亦然。我能想到的最简单的方法是:

 spinner1.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() {
            @Override
            public void onItemSelected(AdapterView<?> parent, View view,
                                                        int position, long id) {
                if (position == spinner2.getSelectedItemPosition()){
                     //UNDO OR CANCEL THE SELECTION SOMEHOW
                }

            }

            @Override
            public void onNothingSelected(AdapterView<?> parent) {

            }
        });

问题是如何撤消对前一个微调器的选择？是否可以不保留全局变量？谢谢

Answer 1

您可以使用堆栈来实现此目的:

Stack<Integer> stack = new Stack<>();
boolean programaticalyChange = false;
@Override
protected void onCreate(Bundle savedInstanceState) {
//Here initialize all you need
spinner.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() {
        @Override
        public void onItemSelected(AdapterView<?> parent, View view, int position, long id) {
            if (programaticalyChange) {
                programaticalyChange = false;
            } else {
                stack.push(position);
            }
        }

        @Override
        public void onNothingSelected(AdapterView<?> parent) {}
    });
//I use button to pop stack back, but you can use any callback from second spinner
btn.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            programaticalyChange = true;
            if (queue.isEmpty()) {
                //no backstack available
                sp.setSelection(0);
            } else {
                if (queue.peek() == sp.getSelectedItemPosition()) queue.pop();
                sp.setSelection(queue.pop());
            }
        }
    });
}