根据教授如何创建弹出窗口的YouTube教程视频(https://www.youtube.com/watch?v=wxqgtEewdfo),我想知道如何使用触摸事件而不是后退按钮来关闭弹出窗口。 ..
这是MainActivity:
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
RelativeLayout relative = (RelativeLayout)findViewById(R.id.relativeTest);
Button button = (Button)findViewById(R.id.button);
button.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
LayoutInflater layoutInflater = (LayoutInflater) getApplicationContext().
getSystemService(LAYOUT_INFLATER_SERVICE);
ViewGroup container = (ViewGroup) layoutInflater.inflate(R.layout.myLayout, null);
PopupWindow popupWindow = new PopupWindow(container, AbsListView.LayoutParams.MATCH_PARENT,
AbsListView.LayoutParams.MATCH_PARENT, true);
popupWindow.showAtLocation(relative, Gravity.CENTER, 0, 0);
container.setOnTouchListener(new View.OnTouchListener() {
@Override
public boolean onTouch(View v, MotionEvent event) {
popupWindow.dismiss();
return true;
}
});
}
});
}
@Override
public void onBackPressed() {
// I want the back button to be disabled for both MainActivity and the
// popup window.
}
...我应该将onBackPressed()置于其他地方,还是可能的?
提前致谢。
答案 0 :(得分:0)
将onBackPressed留空。删除super.OnBackPressed行
答案 1 :(得分:0)
试试这个: -
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if (Integer.parseInt(android.os.Build.VERSION.SDK) > 5
&& keyCode == KeyEvent.KEYCODE_BACK) {
Log.d("CDA", "onKeyDown Called");
onBackPressed();
return true;
}
return super.onKeyDown(keyCode, event);
}
答案 2 :(得分:0)
好的,我现在想出来了(归功于Filip,YouTube视频的上传者)...问题在于我将PopupWindow的最后一个参数设置为true,并将其设置为true,并且我得到了摆脱它如下:
popupWindow = new PopupWindow(container, AbsListView.LayoutParams.MATCH_PARENT,
AbsListView.LayoutParams.MATCH_PARENT);
...弹出窗口现在只能通过触摸事件解除。