Android sqlite选择不工作

我正在使用sqllite。我成功地创建了sql数据库，并且我还可以插入一些值并在listview中显示它。现在我想在ServerID喜欢我的值的地方编写返回标题的函数 我编写了这个函数，但这个函数没有返回计数器

public class DatabaseHandler extends SQLiteOpenHelper {
private static final int DATABASE_VERSION = 1;
private static final String DATABASE_NAME = "lv_db4";
private static final String TABLE_CONTACTS = "CardTable1";
public static final String KEY_ID = "id";

private static final String KEY_Tittle = "title";
private static final String KEY_Description = "description";
private static final String KEY_Price = "price";
private static final String KEY_Counter = "counter";

private static final String KEY_ServerId = "serverid";
private static final String KEY_Image = "image";
private final ArrayList<Contact> contact_list = new ArrayList<Contact>();
public static SQLiteDatabase db;


public DatabaseHandler(Context context) {
    super(context, DATABASE_NAME, null, DATABASE_VERSION);
}

@Override
public void onCreate(SQLiteDatabase db) {
    String CREATE_CONTACTS_TABLE = "CREATE TABLE " + TABLE_CONTACTS + "("
            + KEY_ID + " INTEGER PRIMARY KEY,"

            + KEY_Tittle + " TEXT,"

            + KEY_Description + " TEXT,"

            + KEY_Price + " TEXT,"

            + KEY_Counter + " TEXT,"

            + KEY_ServerId + " TEXT,"

            + KEY_Image + " TEXT"

            + ");";
    db.execSQL(CREATE_CONTACTS_TABLE);
}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    db.execSQL("DROP TABLE IF EXISTS " + TABLE_CONTACTS);
    onCreate(db);
}

// Adding new contact
public void Add_Contact(Contact contact) {
    db = this.getWritableDatabase();
    ContentValues values = new ContentValues();

    values.put(KEY_Tittle, contact.getTitle());
    values.put(KEY_Description, contact.getDescription());
    values.put(KEY_Price, contact.getPrice());

    values.put(KEY_Counter, contact.getCounter());

    values.put(KEY_ServerId, contact.getServerId());

    values.put(KEY_Image, contact.getImage());

    db.insert(TABLE_CONTACTS, null, values);

    db.close();

}

public void deleteUser(String userName) {
    db = this.getWritableDatabase();
    try {
        db.delete(TABLE_CONTACTS, "title = ?", new String[] { userName });

    } catch (Exception e) {
        e.printStackTrace();
    } finally {
        db.close();
    }
}

public String GetCounterFromServerID(String value) {



    db = this.getWritableDatabase();

    Cursor cursor = db.rawQuery("select " + KEY_Counter+  " from " + TABLE_CONTACTS
            + " where serverid like '%" + value + "%'", null);


    cursor.close();
    return KEY_Counter;

}

public int Update_Contact(Contact contact) {
    db = this.getWritableDatabase();

    ContentValues values = new ContentValues();
    values.put(KEY_Counter, contact.getCounter());

    // updating row
    return db.update(TABLE_CONTACTS, values, KEY_ID + " = ?",
            new String[] { String.valueOf(contact.getID()) });
}

// Getting single contact
Contact Get_Contact(int id) {
    SQLiteDatabase db = this.getReadableDatabase();

    Cursor cursor = db.query(TABLE_CONTACTS, new String[] { KEY_ID,
            KEY_Tittle, KEY_Description, KEY_Price, KEY_Counter,
            KEY_ServerId, KEY_Image }, KEY_ID + "=?",
            new String[] { String.valueOf(id) }, null, null, null);
    if (cursor != null)
        cursor.moveToFirst();

    Contact contact = new Contact(cursor.getString(0), cursor.getString(1),
            cursor.getString(2), cursor.getString(3), cursor.getString(4),
            cursor.getString(5));

    cursor.close();
    db.close();

    return contact;
}

public boolean DublicateValues(String S_title) {
    db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery("select * from " + TABLE_CONTACTS
            + " where title like '%" + S_title + "%'", null);
    boolean exists = (cursor.getCount() > 0);

    cursor.close();
    return exists;
}

日志结果是唯一的计数器 我做错了什么？如果有人知道解决方案，请帮助我这是错误的

return KEY_Counter;

因为这是您的字符串

“counter”

。你应使用：

String result = "no data found";
if (cursor != null && cursor.moveToFirst())
    result = cursor.getString(cursor.getColumnIndex(KEY_Counter));
cursor.close();
return result;

改为

Cursor cursor = db.rawQuery("select " + KEY_Counter+  " from " + TABLE_CONTACTS
        + " where serverid like '%" + value + "%'", null);
if(cursor.moveToFirst())
String counter = cursor.getString(5); 
cursor.close();

现在返回

计数器

，而不是

键计数器

，您可以这样做

   public String GetCounterFromServerID(String value)
   {
     Cursor cursor = db.rawQuery("SELECT " + KEY_Counter + " FROM " + TABLE_CONTACTS + " WHERE serverid like '%" + value + "%'", null);

     if (!cursor.moveToFirst())
     {
         return null;
     }
     if(-1 == cursor.getColumnIndex(KEY_Counter)) 
     {
         return null;
     }
     String counter = cursor.getString(cursor.getColumnIndex(KEY_Counter));
     cursor.close();
     return counter;
  }

---

使用

光标

获取值<代码>返回游标。getString（5）我不明白您的意思，先生。我可以在哪里添加此行？你能告诉我@Raghunandanthank you吗？我更改了代码，但你的代码包含错误。这是我的日志错误@user3804530无法读取第0行第5列。这本身应该会给出提示。有什么问题吗query@user3804530并确保

键计数器

位于第5列

   public String GetCounterFromServerID(String value)
   {
     Cursor cursor = db.rawQuery("SELECT " + KEY_Counter + " FROM " + TABLE_CONTACTS + " WHERE serverid like '%" + value + "%'", null);

     if (!cursor.moveToFirst())
     {
         return null;
     }
     if(-1 == cursor.getColumnIndex(KEY_Counter)) 
     {
         return null;
     }
     String counter = cursor.getString(cursor.getColumnIndex(KEY_Counter));
     cursor.close();
     return counter;
  }