Question

我正在尝试使用listfragment来显示20个项目。因为它只显示列表中所有二十个项目的列表中的最后一项。我一直在看listfragments的例子，我不知道出了什么问题。

这是来自扩展ListFragment的CourseFragment类：

private List<Course> mCourses;

public CourseFragment() {
}

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    mCourses = Schedule.get(getActivity()).getSchedule();
    CourseArrayAdapter<Course> mAdapter = new CourseArrayAdapter<>(getContext(), android.R.layout.simple_list_item_1, mCourses);
    setListAdapter(mAdapter);

    for (int i = 0; i < 20; i++ ) {
        Log.d("DEBUGTAG", "Course: " + mCourses.get(i));
    }

}

这是来自CourseArrayAdapter：

public CourseArrayAdapter(Context context, int resource, List<T> objects) {
    super(context, resource, objects);
}

@Override
public View getView(int position, View convertView, ViewGroup parent) {
    Log.d("DEBUG", "getView " + position);
    LayoutInflater inflater = LayoutInflater.from(getContext());
    View mLayout;
    if (convertView == null) {
        mLayout = inflater.inflate(R.layout.view_course, parent, false);
    } else {
        mLayout = convertView;
    }
    T mCourse = getItem(position);
    TextView mCourseTextView = (TextView) mLayout.findViewById(R.id.course_textview);
    mCourseTextView.setText(mCourse.toString());

    return mLayout;
}

MainActivity的onCreate：

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    FragmentManager mFragManager = getFragmentManager();
    Fragment mFragment = mFragManager.findFragmentById(R.id.main_activity_framelayout);
    if(mFragment == null) {
        mFragment = new CourseFragment();
        mFragManager.beginTransaction().add(R.id.main_activity_framelayout, mFragment).commit();
    }

}

以下是Course类的实例变量和构造函数：

private static String sCourseName;
private static int sCourseNum;
private static String sInstructor;
private static String sDescription;

public Course(String mCourseName, int mCourseNum, String mInstructor, String mDescription) {
    sCourseName = mCourseName;
    sCourseNum = mCourseNum;
    sInstructor = mInstructor;
    sDescription = mDescription;
}

如果需要任何其他代码来解决此问题，请告知我们。非常感谢你的帮助！

Answer 1

现在请试试这个mCourses会有数据。我认为不会重复

private List<Course> mCourses;
public CourseArrayAdapter(Context context, int resource, List<T> objects) {
super(context, resource, objects);
mCourses=objects;
}

@Override
public View getView(int position, View convertView, ViewGroup parent) {
Log.d("DEBUG", "getView " + position);
LayoutInflater inflater = LayoutInflater.from(getContext());
View mLayout;
if (convertView == null) {
    mLayout = inflater.inflate(R.layout.view_course, parent, false);
} else {
    mLayout = convertView;
}
  // T mCourse = getItem(position);
TextView mCourseTextView = (TextView) mLayout.findViewById(R.id.course_textview);
mCourseTextView.setText(mCourses.get(position).toString());

return mLayout;
}

Android ListFragment仅显示重复的最后一项

1 个答案: