Question

尝试在我的Android应用程序中创建一个sqlite数据库。表创建和值插入工作完美，但是当我尝试从db中获取（选择）值时，使用

android.database.sqlite.SQLiteException: no such column: rowId (code 1): , while compiling: SELECT rowId, firstName, lastName, photoPrefix;, photoSuffix;, hasPhoto FROM friends WHERE  rowId = ?

这是我的SQLiteOpenHelper继承类

的属性

private static final int DATABASE_VERSION = 1;
private static final String DATABASE_NAME = "GroupDB";
private static final String TABLE_GROUPS = "groups";

private static final String KEY_ID = "rowId";
private static final String KEY_NAME = "name";
private static final String KEY_FRIENDS = "friends";
private static Context context;
private static final String[] COLUMNSGroup = { KEY_ID, KEY_NAME,
        KEY_FRIENDS };

private static final String TABLE_FRIENDS = "friends";
private static final String KEY_FIRST_NAME = "firstName";
private static final String KEY_LAST_NAME = "lastName";
private static final String KEY_PHOTO_PREFIX = "photoPrefix;";
private static final String KEY_PHOTO_SUFFIX = "photoSuffix;";
private static final String KEY_HAS_PHOTO = "hasPhoto";
private static final String[] COLUMNSFriends = { KEY_ID, KEY_FIRST_NAME,
        KEY_LAST_NAME, KEY_PHOTO_PREFIX, KEY_PHOTO_SUFFIX, KEY_HAS_PHOTO };

onCreate方法：

@Override
public void onCreate(SQLiteDatabase db) {
    String CREATE_GROUP_TABLE = "CREATE TABLE groups ( "
            + "rowId INTEGER PRIMARY KEY AUTOINCREMENT, " + "name TEXT, "
            + "friends TEXT )";

    db.execSQL(CREATE_GROUP_TABLE);
    String CREATE_FRIEND_TABLE = "CREATE TABLE friends ( " + "rowId TEXT, "
            + "firstName TEXT, " + "lastName TEXT, " + "photoPrefix TEXT, "
            + "photoSuffix TEXT, " + "hasPhoto INTEGER )";
    db.execSQL(CREATE_FRIEND_TABLE);
    Log.d("DB", CREATE_GROUP_TABLE);
    Log.d("DB", CREATE_FRIEND_TABLE);
}

addFriend方法也正常工作：

public void addFriend(Friends friend) {
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues values = new ContentValues();
    values.put(KEY_ID, friend.id);
    values.put(KEY_FIRST_NAME, friend.firstName);
    values.put(KEY_LAST_NAME, friend.LastName);
    values.put(KEY_PHOTO_PREFIX, friend.photoPrefix);
    values.put(KEY_PHOTO_SUFFIX, friend.photoSuffix);
    values.put(KEY_HAS_PHOTO, friend.hasPhoto ? 1 : 0); 

    db.insert(TABLE_FRIENDS, null,values); 
    }
    db.close();
}

getFriend方法给出了错误

public Friends getFriend(String id) {

    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.query(TABLE_FRIENDS, //THIS 
            COLUMNSFriends, //LINE
            " rowId = ?",  //GIVES
            new String[] { id }, null, null, null, null); //EXCEPTION
    if (cursor != null)
        cursor.moveToFirst();
    Friends friend = new Friends();
    friend.id = cursor.getString(0);
    friend.firstName = cursor.getString(1);
    friend.LastName = cursor.getString(2);
    friend.photoPrefix = cursor.getString(3);
    friend.photoSuffix = cursor.getString(4);
    friend.hasPhoto = cursor.getString(5).equals("1") ? true : false;
    return friend;
}

Answer 1

我不知道Android SQLite API - 但是在其他sqlite APIS上你需要明确地指出你的数据库更改 - 所以在插入不执行后执行关闭，它可能会回滚。

Answer 2

我错过了这部分，改变了“photoSuffix;”;到“photoSuffix”;和“photoPrefix;”;解决这个问题。谢谢Raghunandan！

Answer 3

使用此查询：

Cursor cursor = db.query(TABLE_NAME,new String[]{ID,FIRST_NAME,LAST_NAME,EMAIL_ID,PASSWORD,PH_NUMB,POST_ADDRESS},null,null,null,null,FIRST_NAME);

没有这样的列：rowId（代码1）在android中编译sqlite时

3 个答案: