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我希望用户多次更改按钮的文本。为此，他长按该按钮。这是代码：

@Override
public void onCreate(Bundle savedInstanceState) {

//blah blah

    final AlertDialog.Builder alert = new AlertDialog.Builder(this);

    alert.setMessage("Nueva Categoria:");

    // Seting an EditText view to get user input 
    final EditText input = new EditText(this);
    alert.setView(input);

    alert.setPositiveButton("Ok", new DialogInterface.OnClickListener() {
    public void onClick(DialogInterface dialog, int whichButton) {
        Button esteBoton = (Button) findViewById(R.id.button1);
        String newCateg = input.getText().toString();
        esteBoton.setText(newCateg);
      }
    });       


    Button button = (Button) findViewById(R.id.button1);
    button.setOnLongClickListener(new View.OnLongClickListener() {
        public boolean onLongClick(View v) {
            alert.show();               
            return true;
        }
    });
}

行。当我在 Eclipse 的设备模拟器中运行此代码时，如果这是我第一次在警报对话框中为按钮 1 输入文本，则没有问题，但如果我第二次尝试输入代码，应用程序将崩溃。我不是Java专家，但我认为这是由“输入”的“最终”属性引起的，一旦确定，我就无法更改它的值。我该如何解决？代码很简单，我想保持这种方式。

4

2 回答 2

1

尝试删除 onCreate 中的构建器部分并将其移动到 onLongClickListener

Button button;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    setContentView(R.layout.symptoms);
    button = (Button) findViewById(R.id.btDone);

    // final Dialog alert = builder.create();

    button.setOnLongClickListener(new View.OnLongClickListener() {
        public boolean onLongClick(View v) {

            // Declare your builder here - 
            final AlertDialog.Builder builder = new AlertDialog.Builder(
                    YOURACTIVITY.this);
            builder.setMessage("Nueva Categoria:");
            // Seting an EditText view to get user input
            final EditText input = new EditText(YOURACTIVITY.this);
            builder.setView(input);
            builder.setPositiveButton("Ok",
                    new DialogInterface.OnClickListener() {
                        public void onClick(DialogInterface dialog,
                                int whichButton) {
                            String newCateg = input.getText().toString();
                            button.setText(newCateg);
                        }
                    });

            builder.show();
            return true;
        }
    });
}

试试这个，看看这是否有效。

于 2013-03-05T06:36:45.477 回答
0

试试下面的代码：

 public class MainActivity extends Activity {
           Button button;
           Context context;
     @Override
 public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    context = MainActivity.this;
    setContentView(R.layout.activity_main);
    button = (Button) findViewById(R.id.button1);
    button.setOnLongClickListener(new View.OnLongClickListener() {
        public boolean onLongClick(View v) {
            final AlertDialog.Builder alert = new AlertDialog.Builder(
                    context);
            alert.setMessage("Nueva Categoria:");
            // Seting an EditText view to get user input
            final EditText input = new EditText(context);
            alert.setView(input);
            alert.setPositiveButton("Ok",
                    new DialogInterface.OnClickListener() {
                        public void onClick(DialogInterface dialog,
                                int whichButton) {
                            String newCateg = input.getText().toString();
                            button.setText(newCateg);
                        }
                    });
            AlertDialog build = alert.create();
            build.show();
            return true;
        }
    });
 }
  }

在按钮的 onLongClickListener 中定义您的对话框。查看代码，它现在工作得很棒。

于 2013-03-05T05:30:18.757 回答