Question

我使用SQLite创建了一个数据库，我似乎正确地创建了数据库，空间和所有，但是当我运行应用程序时，我得到的异常是它找不到列id：“_ id”

请查看我的代码，让我知道我做错了什么。

public static final String KEY_ROW_ID = "_id";
public static final String KEY_NAME = "persons_name";
public static final String RATE = "rate_it";

private static final String DATABASE_NAME = "iDRateIt";
private static final String DATABASE_TABLE = "tblPeople";
private static final int DATABASE_VERSION = 1;

private dbHelper ourHelper;
private final Context ourContext;
private SQLiteDatabase ourDatabase;

private static class dbHelper extends SQLiteOpenHelper {

    public dbHelper(Context context) {
        super(context, DATABASE_NAME, null, DATABASE_VERSION);
        // TODO Auto-generated constructor stub
    }

    @Override
    public void onCreate(SQLiteDatabase db) {
        // TODO Auto-generated method stub
        db.execSQL("CREATE TABLE " + DATABASE_TABLE + " ( " +
                KEY_ROW_ID + " INTEGER PRIMARY KEY AUTOINCREMENT, " +
                KEY_NAME + " TEXT NOT NULL, " +
                RATE + " TEXT NOT NULL);"

                );
    }

    @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
        // TODO Auto-generated method stub
        db.execSQL("DROP TABLE IF EXISTS " + DATABASE_TABLE);
        onCreate(db);
    }

}

//Constructor
public RateIt (Context c) {
    ourContext = c;
}

public RateIt open() throws SQLException
{
    ourHelper = new dbHelper(ourContext);
    ourDatabase = ourHelper.getWritableDatabase();
    return this;
}
//Close Database
public void close()
{
    ourHelper.close();
}

//Insert Method
public long createEntry(String name, String rate) {
    // TODO Auto-generated method stub
    ContentValues cv = new ContentValues();

    cv.put(KEY_NAME, name);
    cv.put(RATE, rate);
    return ourDatabase.insert(DATABASE_TABLE, null, cv);


}

public String getData() {
    // TODO Auto-generated method stub
    String[] columns = new String[]{ KEY_ROW_ID, KEY_NAME, RATE};
    Cursor c = ourDatabase.query(DATABASE_TABLE, columns, null, null, null, null, null);

    String result = "";

    int iRow = c.getColumnIndex(KEY_ROW_ID);
    int iName = c.getColumnIndex(KEY_NAME);
    int iRate = c.getColumnIndex(RATE);

    for (c.moveToFirst(); !c.isAfterLast(); c.moveToNext())
    {
        result = result + c.getString(iRow) + " " + c.getString(iName) 
                + " " + c.getString(iRate) + "\n";

    }

    return result;

}

}

Answer 1

你确定它正确创造吗？因为

db.execSQL("CREATE TABLE " + DATABASE_TABLE + " ( " +
        KEY_ROW_ID + " INTEGER PRIMARY KEY AUTOINCREMENT, " +
        KEY_NAME + " TEXT NOT NULL, " +
        RATE + " TEXT NOT NULL);"

似乎有一个分散的分号，应该是：

db.execSQL("CREATE TABLE " + DATABASE_TABLE + " ( " +
        KEY_ROW_ID + " INTEGER PRIMARY KEY AUTOINCREMENT, " +
        KEY_NAME + " TEXT NOT NULL, " +
        RATE + " TEXT NOT NULL)"

Answer 2

看起来

存在问题

c.getString(iRow)

从代码中

将_id列定义为整数

db.execSQL("CREATE TABLE " + DATABASE_TABLE + " ( " +
            KEY_ROW_ID + " INTEGER PRIMARY KEY AUTOINCREMENT, "

您需要将此视为整数

int iRow =（c.getLong（c.getColumnIndex（KEY_ROW_ID）））; 或

int iRow = (c.getInteger(c.getColumnIndex(KEY_ROW_ID)));

希望这有帮助。

此外，您已定义费率

RATE + " TEXT NOT NULL

作为文本字段，如果你这样做只是为了测试目的，但良好的数据库设计会将其定义为某种类型的数字字段，即Integer，Long。

度过美好的一天。

SQLite Exception没有列id找到Android

2 个答案: