我有以下代码创建一个包含EditText的PopupWind:
lbs.setOnTouchListener(new OnTouchListener(){
@Override
public boolean onTouch(View v, MotionEvent arg1) {
int pWidth = 100;
int pHeight = 80;
int vHeight = mView.getHeight();
int[] location = new int[2];
v.getLocationOnScreen(location);
final View view = inflater.inflate(R.layout.list_popup, null, false);
final PopupWindow pw = new PopupWindow(view, pWidth, pHeight, false);
pw.setTouchable(true);
//pw.setFocusable(true);
pw.setOutsideTouchable(true);
pw.setBackgroundDrawable(new BitmapDrawable());
pw.setContentView(view);
pw.showAtLocation(v, Gravity.NO_GRAVITY, location[0]-(pWidth/4), location[1]+vHeight);
//final LinearLayout layout = (LinearLayout)view.findViewById(R.id.PopupLayout);
final EditText input = (EditText)view.findViewById(R.id.Input);
input.setOnFocusChangeListener(new View.OnFocusChangeListener() {
@Override
public void onFocusChange(View v, boolean hasFocus) {
Log.i("Focus", "Focus Changed");
/*
if (hasFocus) {
InputMethodManager inputMgr = (InputMethodManager)myContext.getSystemService(Context.INPUT_METHOD_SERVICE);
inputMgr.toggleSoftInput(InputMethodManager.SHOW_FORCED, 0);
inputMgr.showSoftInput(v, InputMethodManager.SHOW_IMPLICIT);
}
*/
}
});
input.setText(lbs.getText().toString());
input.requestFocus();
pw.setOnDismissListener(new OnDismissListener(){
@Override
public void onDismiss() {
parentActivity.changeWeight(getId, Double.parseDouble(input.getText().toString()));
Log.i("View Visibility", "" + view.getVisibility());
}
});
pw.setTouchInterceptor(new View.OnTouchListener() {
@Override
public boolean onTouch(View v, MotionEvent event) {
if (event.getAction() == MotionEvent.ACTION_OUTSIDE) {
Log.i("Background", "Back Touched");
pw.dismiss();
return true;
}
return false;
}
});
return true;
}
});
PopupWindows有一个setBackGroundDrawable,当用户触摸窗口外时,该框应自动关闭。我知道这个盒子正在解雇,因为当我在盒子外面点击时我的OnDismiss方法运行,但实际窗口不会消失,直到我在盒子外面触摸两次,因为onDismiss运行两次,我不能。我在这里缺少什么?
编辑:我稍微更新了我的代码。如果我不把注意力集中在弹出窗口上,除了我无法专注于我的编辑文本之外,一切都很完美。如果我在弹出窗口中执行setFocusable,则edittext不会立即获得焦点,我必须双击才能删除它。
答案 0 :(得分:3)
我认为您可能会打开多个弹出窗口。 onTouch方法将被调用至少用于触摸和触摸,并且可能还有一些触摸移动。尝试检查(arg1.getAction()== MotionEvent.ACTION_UP)并仅显示窗口。
答案 1 :(得分:1)
您的弹出窗口未获取父活动的上下文。 尝试这样做而不是“最终PopupWindow pw = new PopupWindow(view,pWidth,pHeight,false);”
final PopupWindow pw = new PopupWindow(getApplicationContext());
pw.setContentView(view);
pw.setHeight(pHeight);
pw.setWidth(pWidth);
pw.setFocusable(false);