Question

我在API 23上的自定义对话框中面临一个问题，它正在崩溃。

final Dialog dialog = new Dialog(getActivity());
dialog.getWindow().setType(WindowManager.LayoutParams.TYPE_SYSTEM_ALERT);
dialog.requestWindowFeature(Window.FEATURE_NO_TITLE);
dialog.setContentView(R.layout.progress_dialog);
dialog.setCancelable(false);
dialog.show();

崩溃报告：

android.view.WindowManager$BadTokenException: Unable to add window android.view.ViewRootImpl$W@3387043 -- permission denied for this window type
at android.view.ViewRootImpl.setView(ViewRootImpl.java:598)
at android.view.WindowManagerGlobal.addView(WindowManagerGlobal.java:310)
at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:86)
at android.app.Dialog.show(Dialog.java:319)
at com.example.hello.fragment.ExploreFragment.exploreBeers(ExploreFragment.java:255)
at com.example.hello.fragment.ExploreFragment$4.onEditorAction(ExploreFragment.java:203)
at android.widget.TextView.onEditorAction(TextView.java:4773)
at com.android.internal.widget.EditableInputConnection.performEditorAction(EditableInputConnection.java:139)
at com.android.internal.view.IInputConnectionWrapper.executeMessage(IInputConnectionWrapper.java:304)
at com.android.internal.view.IInputConnectionWrapper$MyHandler.handleMessage(IInputConnectionWrapper.java:78)
at android.os.Handler.dispatchMessage(Handler.java:102)
at android.os.Looper.loop(Looper.java:152)
at android.app.ActivityThread.main(ActivityThread.java:5497)
at java.lang.reflect.Method.invoke(Native Method)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:726)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:616)

清单许可：

<uses-permission android:name="android.permission.SYSTEM_ALERT_WINDOW" />

根本没有工作我有点困惑，请通过我的帖子并建议我解决一些问题。

Answer 1

删除。dialog.getWindow（）的setType（WindowManager.LayoutParams.TYPE_SYSTEM_ALERT）; 并在您的应用程序中尝试。或者如果你想在第三方应用程序上显示它

Answer 2

查看我的answer on runtime permissions，

在本回答中，您可以将代码包含在访问联系人方法中。

在任何活动中实施权限的示例。

SampleActivity.java

    public class SampleActivity extends AppCompatActivity{
        private final int PERMISSION_CODE = 1;
        Button button;
        @override
        onCreate(Bundle savedInstance){
            super.onCreate(savedInstance);
            setContentView(R.layout.your_layout);
            button = (Button)findViewById(R.id.button);
            button.setOnClickListener(new OnClickListener(){
                @override
                public void onClick(View view){
                    requestPermissionAndContinue();
                }
            });
            //remaining code to continue using the app
            //your actual code should also be in this same class
        }

        private void requestPermissionAndContinue(){
            if(ContextCompat.checkSelfPermission(this, Manifest.permission.SYSTEM_ALERT_WINDOW) != PackageManager.PERMISSION_GRANTED){
                if(ActivityCompat.shouldShowRequestPermissionRationale(this, Manifest.permission.SYSTEM_ALERT_WINDOW)){
                    Log.e(TAG, "permission denied, show dialog");
                }else{
                    ActivityCompat.requestPermissions(this, new String[]{Manifest.permission.SYSTEM_ALERT_WINDOW}, PERMISSION_CODE);
                }
            }else{
                showAlert();
            }
        }

        private void showAlert(){
            //your code to show alert dialog
            final Dialog dialog = new Dialog(getActivity());
  dialog.getWindow().setType(WindowManager.LayoutParams.TYPE_SYSTEM_ALERT);
dialog.requestWindowFeature(Window.FEATURE_NO_TITLE);
dialog.setContentView(R.layout.progress_dialog);
dialog.setCancelable(false);
dialog.show();
        }

        @Override
        public void onRequestPermissionsResult(int requestCode, @NonNull String[] permissions, @NonNull int[] grantResults) {
            if(grantResults.length > 0 && permissions[0]==PackageManager.PERMISSION_GRANTED){
                showAlert();
            }else{
                 //redirect to settings page or ask permission again
            }
        }
    }

为什么自定义对话框无法在API 23上运行？

2 个答案: