Question

我已经设置了一个列表，当我触摸列表中的一个项目时，我希望该应用程序生成一个AlertDialog。但是，当我触摸一个项目时，应用程序崩溃，我现在知道为什么。这是我的代码。

DATA = new ArrayList<Student>();
    adapter1 = new StudentAdapter(this, R.layout.studentitemlayout, DATA);
    listview01 = (ListView) findViewById(R.id.ListView01);
    listview01.setAdapter(adapter1);

    listview01.setOnItemClickListener(new AdapterView.OnItemClickListener() {
        public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
            Log.i(DEBUG_TAG, "You clicked " + position + " student");
            AlertDialogFragment fragment = new AlertDialogFragment();
            fragment.onCreateDialog(savedInstanceState).show();
        }
    });

和我的片段

public class AlertDialogFragment extends DialogFragment {

private static final String DEBUG_TAG = "MyStudentsData";

public Dialog onCreateDialog(Bundle savedInstanceState) {
    // Use the Builder class for convenient dialog construction
    AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
    builder.setMessage("New Lesson!")
            .setPositiveButton("Yes", new DialogInterface.OnClickListener() {
                public void onClick(DialogInterface dialog, int id) {
                    Log.i(DEBUG_TAG , "YES");
                }
            })
            .setNegativeButton("No", new DialogInterface.OnClickListener() {
                public void onClick(DialogInterface dialog, int id) {
                    Log.i(DEBUG_TAG , "NO");
                }
            });
    // Create the AlertDialog object and return it
    return builder.create();
}

}

一个最终问题为什么这段代码失败并让我在应用程序上崩溃？

/*newLessonAlertDialog fragment = new newLessonAlertDialog();
            fragment.show(getFragmentManager(),DEBUG_TAG); */
            AlertDialog alert = new AlertDialog.Builder(getApplicationContext()).create();
            alert.setTitle("nikos");
            alert.show();

Answer 1

使用show调用它。无需致电onCreateDialog

fragment.show(getFragmentManager(), "Some tag");

Answer 2

您正在使用DialogFragment，因此您应该调用fragment.show（）而不是onCreateDialog方法。

应用程序在使用AlertDialogs时崩溃

2 个答案: