Android长按自定义键盘弹出键盘

Question

我有自定义Android键盘:

    public class CustomKeyboard extends Keyboard{...}  

    public class CustomKeyboardView extends KeyboardView{...}

    public class CustomKeyboardIME extends InputMethodService implements KeyboardView.OnKeyboardActionListener{...}  

在一些按键,我有popupKeyboard和popupCharacters:

<Key android:codes="144" android:keyLabel="0" android:popupKeyboard="@xml/key_popup" android:popupCharacters=")" android:keyEdgeFlags="right"/>

XML/key_popup.xml:

<Keyboard xmlns:android="http://schemas.android.com/apk/res/android"
      android:keyWidth="10%p"
      android:horizontalGap="0px"
      android:verticalGap="0px"
      android:keyHeight="@dimen/key_height" >
</Keyboard>

但是当我长按"0"键弹出")"显示时,它会一直保持在那里直到我按下"X"按钮或")"字符.它看起来像这样:

而且我希望它只在我按住手指时打开.像三星或HTC键盘上的东西:

有谁可以帮助我吗？

编辑至少可以改变这个弹出窗口的外观吗？我希望它具有与我制作的整个键盘相同的背景和键

Answer 1

您可以使用PopupWindow类并使用自定义布局填充它.

View custom = LayoutInflater.from(context)
    .inflate(R.layout.your_layout, new FrameLayout(context));
PopupWindow popup = new PopupWindow(context);
popup.setContentView(custom);

并且长按

//Get x,y based on the touch position
//Get width, height based on your layout
if(popup.isShowing()){
    popup.update(x, y, width, height);
} else {
    popup.setWidth(width);
    popup.setHeight(height);
    popup.showAtLocation(yourKeyboardView, Gravity.NO_GRAVITY, x, y);
}

点击弹出窗口即可解除它

popup.dismiss();