android - 启动时选择的 Android Spinner 设置项目

Question

我的活动中有一个 Spinner，其中包含一个字符串列表，它的声明如下：

String[] notificationStatus = new String[statuses.size()];
        for (int i =0; i<statuses.size();i++){
            notificationStatus[i]=statuses.get(i);
        }
        ArrayAdapter<String> aa = new ArrayAdapter<String> (
                this, android.R.layout.simple_spinner_item, notificationStatus );

        aa.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item);
        spinner.setAdapter(aa);
        spinner.setPrompt(getString(R.string.chooseType));
        spinner.setOnItemSelectedListener(new OnItemSelectedListener() {
            public void onItemSelected(AdapterView<?> parentView, View selectedItemView, int position, long id) {
                posType=position;
            }

            public void onNothingSelected(AdapterView<?> parentView) {
                // your code here
            }

            });

现在，当我开始这个活动时，我在按钮上得到了 notificationStatus 数组的第一项（微调器的位置 0），有没有办法开始一个活动并将例如数组的第三项放在 2 的位置？

Answer 1

Actually i have found out how to do this. First i create a list from the array:

List<String> statusCheck = new ArrayList<String>();
        statusCheck = Arrays.asList(notificationStatus);

Next i check where on the list is String i am looking for and put set selection on it:

spinner2.setSelection(statusCheck.indexOf(status));