java - android studio - 如何将微调器添加到服务中的弹出窗口

Question

我这里遇到了一些问题!我的代码显示了我的服务中的弹出窗口，并且在该弹出窗口内有一个微调器，当我单击微调器时，错误开始向我展示:'(这是使用的代码:

            final LayoutInflater inflater = (LayoutInflater) getSystemService(Context.LAYOUT_INFLATER_SERVICE);
            final PopupWindow pw = new PopupWindow(inflater.inflate(R.layout.popup_window, null, false), ViewGroup.LayoutParams.WRAP_CONTENT, ViewGroup.LayoutParams.WRAP_CONTENT, true);
            final Spinner spn_ = (Spinner)pw.getContentView().findViewById(R.id.spinner3);
            final String[] options_alma_ = {"Hello","HI"};
            final ArrayAdapter<String> stringArrayAdapter_ = new ArrayAdapter<String>(AlMA_Service.this, android.R.layout.simple_spinner_dropdown_item, options_alma_);
            final Button btn = (Button)pw.getContentView().findViewById(R.id.button10);
            btn.setText("Choose Hello");
            btn.setOnClickListener(new View.OnClickListener() {
                @Override
                public void onClick(View v) {
                    Log.e("Clicked", "Here Before Dismiss !!");
                    pw.dismiss();
                }
            });
            spn_.setAdapter(stringArrayAdapter_);
            spn_.setSelection(stringArrayAdapter_.getPosition(Locale.getDefault().getDisplayName()));
            spn_.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() {
                @Override
                public void onItemSelected(AdapterView<?> parent, View view, int position, long id) {
                    try {
                        btn.setText("Choose " + stringArrayAdapter_.getItem(position));
                    } catch (NullPointerException e) {
                        e.printStackTrace();
                        btn.setText("Choose Hello" );
                        }
                }
                @Override
                public void onNothingSelected(AdapterView<?> parent) {
                }
            });
            pw.showAsDropDown(mView);

有什么帮助吗？

Answer 1

您的代码的问题是您无法可靠地将当前 Activity 的 Activity 传递到弹出窗口 会有一些不可靠地传递 Activity 的方式，但我不推荐它

你可以尝试这个而不是弹出菜单

windowManager2 = (WindowManager)getSystemService(WINDOW_SERVICE);
 LayoutInflater layoutInflater=(LayoutInflater)getSystemService(Context.LAYOUT_INFLATER_SERVICE); 
View view=layoutInflater.inflate(R.layout.popupmenu, null); 
params=new WindowManager.LayoutParams( WindowManager.LayoutParams.WRAP_CONTENT, WindowManager.LayoutParams.WRAP_CONTENT, WindowManager.LayoutParams.TYPE_PHONE, WindowManager.LayoutParams.FLAG_NOT_FOCUSABLE, PixelFormat.TRANSLUCENT);
 params.gravity=Gravity.CENTER|Gravity.CENTER;
 params.x=0; 
params.y=0; 
windowManager2.addView(view, params);