Question

我在以下代码行中获得空指针异常

attendanceList = dbHelper.getAttendance(surName, name);

surName和name都不为null，有size（）＆gt; 0和attendanceList的类型为ArrayList，这里是getAttendance函数：

public ArrayList<Attendance> getAttendance(String surName, String name)
 {
    SQLiteDatabase db = this.getReadableDatabase();
    ArrayList<Attendance> attendanceList = new ArrayList<>();

    Cursor cursor = db.rawQuery("SELECT * FROM " + ATTENDANCE_TABLE + " WHERE " + COLUMN_SURNAME + " =? AND "
            + COLUMN_NAME + " =?", new String[] {surName, name});

    if(cursor .moveToFirst())
    {
        while(!cursor.isAfterLast())
        {
            Attendance attendance = new Attendance();

            attendance.setDate(cursor.getString(cursor.getColumnIndex(COLUMN_DATE)));
            attendance.setName(cursor.getString(cursor.getColumnIndex(COLUMN_NAME)));
            attendance.setSurName(cursor.getString(cursor.getColumnIndex(COLUMN_SURNAME)));
            attendance.setState(cursor.getString(cursor.getColumnIndex(COLUMN_STATE)));
            attendance.setTerm(cursor.getInt(cursor.getColumnIndex(COLUMN_TERM)));
            attendance.setSubject(cursor.getString(cursor.getColumnIndex(COLUMN_SUBJECT_NAME)));

            attendanceList.add(attendance);
            cursor.moveToNext();
        }
        cursor.close();
        db.close();
        return attendanceList;
    }
    else
    {
        cursor.close();
        db.close();
        return null;
    }
}

上面的函数没有语法错误，我知道它不应该返回null，因为我使用DB Browser for SQLite看到了ATTENDANCE_TABLE的内容。有人能指出我的代码有什么问题吗？我觉得我写了1 + 1 = 2并收到错误..

Answer 1

试试这个... SQLiteDbHelper是我的数据库类，我的表已定义。 Table_Name是存储我的表名的字符串

cursor = db.rawQuery("SELECT * FROM "+SQLiteDBHelper.TABLE_NAME+" WHERE "+SQLiteDBHelper.COLUMN_SURNAME+"=? AND "+SQLiteDBHelper.COLUMN_NAME+"=?",new String[] {surName,name});

            if (cursor != null) {
                if(cursor.getCount() > 0) {

                 cursor.moveToFirst();
                    //Retrieving User SurName and Name
                    String s_name = cursor.getString(cursor.getColumnIndex(SQLiteDBHelper.COLUMN_SURNAME));
                    String _name= cursor.getString(cursor.getColumnIndex(SQLiteDBHelper.COLUMN_NAME));
                    Toast.makeText(MainActivity.this, "Success", Toast.LENGTH_SHORT).show();

Answer 2

我很确定导致surName的{{1}}或name不是NullpointerException，而是dbHelper。

因此，请检查您是否在此行之后初始化dbHelper

attendanceList = dbHelper.getAttendance(surName, name);

如果是这样，那么将其移到上面，你的崩溃就会得到解决。希望它有所帮助。

SQLite select查询在不应该返回null时返回null

2 个答案: