Question

我一直在尝试让ListView出现在屏幕上，我必须遗漏一些因为它不起作用。代码非常简单，从数据库中取出数据，然后在适配器和列表设置中生成一个字符串。

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.select_singles);
    db = new DbHelper(ViewSinglesAndAdd.this);
    db.open();
    this.all = db.getAllSingle();
    db.close();
    showList = (ListView) findViewById(R.id.listViewSingles);
    if (all == null)
        Toast.makeText(ViewSinglesAndAdd.this, "Empty", Toast.LENGTH_LONG)
                .show();
    if (all != null) {
        String[] select = new String[all.size()];
        for (int i = 0; i < all.size(); i++)
            select[i] = all.elementAt(i).getPerson().toString();
        ArrayAdapter<String> adapter = new ArrayAdapter<String>(
                ViewSinglesAndAdd.this,
                android.R.layout.simple_list_item_multiple_choice, select);
        showList.setChoiceMode(ListView.CHOICE_MODE_MULTIPLE);

        showList.setAdapter(adapter);



    }

Answer 1

如果all代表您的字符串ArrayList，那么您需要更改

   select[i] = all.elementAt(i).getPerson().toString();

到

   select[i] = all.get(i).getPerson();

使用列表视图

1 个答案: