extend button android, xml layout

I have the following class (included into another class)

class RecordButton extends Button {
    boolean mStartRecording = true;

    OnClickListener clicker = new OnClickListener() {
        public void onClick(View v) {
            onRecord(mStartRecording);
            if (mStartRecording) {
                setText("Stop recording");
      开发者_高级运维      } else {
                setText("Start recording");
            }
            mStartRecording = !mStartRecording;
        }
    };

    public RecordButton(Context ctx) {
        super(ctx);
        setText("Start recording");
        setOnClickListener(clicker);
    }
}

The display of the button is made using the following code:

public void onCreate(Bundle icicle) {
    super.onCreate(icicle);

    LinearLayout ll = new LinearLayout(this);
    mRecordButton = new RecordButton(this);
    ll.addView(mRecordButton,
        new LinearLayout.LayoutParams(
            ViewGroup.LayoutParams.WRAP_CONTENT,
            ViewGroup.LayoutParams.WRAP_CONTENT,
            0));
    setContentView(ll);
}

How can I define the Button layout into the .xml file instead of doing it in the java code?

I have tried that:

<AudioRecordTest.test.RecordButton
    android:layout_width="fill_parent"
    android:layout_height="wrap_content"
    android:text="Button"
    android:id="@+id/record" />   

But it is not working...

Many thanks,

Joachim

I understand "(included into another class)" as you have an inner class RecordButton.

Assuming your package is AudioRecordTest.test (which would be a very bad name choice) and your RecordButton class is an inner class of AudioRecord.class, you need to use:

 <view class="AudioRecordTest.test.AudioRecord$RecordButton"

Use the $ sign to separate inner classes. You need to write the qualified name inside quotes. Also, make sure you create your class public static, or it won't be visible.

BTW: any particular reason you create it as an inner class instead of having it separate?