java - 在 toast 数组列表中显示

Question

如何在 toast 中显示我的数组列表..这是我的代码，但它没有显示数组中的完整项目..我的问题是什么？

public static  EditText txt1;
String ag;
public static ArrayList<String> playerList = new ArrayList<String>();
String playerlist[];

/** Called when the activity is first created. 
 * @param Public */
public void onCreate(Bundle savedInstanceState, Object Public) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.screen2);

    // edittext1 or textview1
    txt1 = (EditText) findViewById(R.id.editText1);
    ag = txt1.getText().toString();

    //add more items button
    Button more = (Button) findViewById(R.id.button1);
    more.setOnClickListener(new View.OnClickListener() {
        public void onClick(View view) {
            if (txt1.getText().length() != 0) {
                String ag = "Current Added Players:," + txt1.getText().toString() + txt1.getText().toString();
                playerList.add(ag);
                Toast.makeText(getBaseContext(), ag, Toast.LENGTH_SHORT).show();

                Intent myIntent = new Intent(view.getContext(), Screen2.class);
                myIntent.putExtra("Stringarray", playerList);
                //startActivityForResult(myIntent, 0);
            }
        }
    });
}

Answer 1

如果您的 ArrayList 是字符串或任何包装类对象，那么您可以按照此操作

ArrayList<String> a = new ArrayList<String>();
a.add("Hello");
a.add("World"); //similarly any other add statements
Toast.makeText(getBaseContext(), a+"", Toast.LENGTH_SHORT).show();

不需要转换为数组或迭代列表，因为当列表是 String 或包装类对象时，它们的 toString() 方法将返回值而不是十六进制逻辑地址。

因此，在这种情况下，您将收到一条显示消息

[Hello,World,...other elements what is stored in the arraylist]