获取SQLiteException：没有这样的列

public class MyDBHandler extends SQLiteOpenHelper { 

    private static final int DATABASE_VERSION = 1; 
    private static final String DATABASE_NAME = "winkelskortrijk.db"; 
    private static final String TABLE_USERS = "users"; 
    private static final String TABLE_FAVOURITES = "favorieten"; 

    public static final String COLUMN_ID = "id"; 
    public static final String COLUMN_USERNAME = "username"; 
    public static final String COLUMN_PASSWORD = "password"; 

    public static final String COLUMN_ID2 = "id2"; 
    public static final String COLUMN_NAME = "naam"; 
    public static final String COLUMN_STRAAT = "straat"; 
    public static final String COLUMN_HUISNR = "nr"; 
    public static final String COLUMN_POSTCODE = "postcode"; 
    public static final String COLUMN_DEELGEMEENTE = "deelgemeente"; 
    public static final String COLUMN_GEMEENTE = "gemeente"; 
    public static final String COLUMN_ADRES = "adres"; 

    public MyDBHandler(Context context, String name, 
         SQLiteDatabase.CursorFactory factory, int version) { 
     super(context, DATABASE_NAME, factory, DATABASE_VERSION); 
    } 

    @Override 
    public void onCreate(SQLiteDatabase db) { 

     String CREATE_USERS_TABLE = "CREATE TABLE " + 
       TABLE_USERS + "(" 
       + COLUMN_ID + " INTEGER PRIMARY KEY AUTOINCREMENT," 
       + COLUMN_USERNAME + " TEXT," 
       + COLUMN_PASSWORD + " TEXT" 
       + ")"; 
     db.execSQL(CREATE_USERS_TABLE); 

     String CREATE_FAVOURITES_TABLE = "CREATE TABLE " + 
       TABLE_FAVOURITES + "(" 
       + COLUMN_ID2 + " INTEGER PRIMARY KEY," 
       + COLUMN_NAME + " TEXT," 
       + COLUMN_STRAAT + " TEXT," 
       + COLUMN_HUISNR + " INTEGER," 
       + COLUMN_POSTCODE + " INTEGER," 
       + COLUMN_DEELGEMEENTE + " TEXT," 
       + COLUMN_GEMEENTE + " TEXT," 
       + COLUMN_ADRES + " TEXT," 
       + COLUMN_ID + " INT, " 
       + "FOREIGN KEY(" + COLUMN_ID + ") REFERENCES " 
       + TABLE_USERS + "(id) " + ")"; 
     db.execSQL(CREATE_FAVOURITES_TABLE); 

    } 

    @Override 
    public void onUpgrade(SQLiteDatabase db, int oldVersion, 
          int newVersion) { 

    } 


    public User findUser(User user) { 

     String selectQuery = "SELECT * FROM " + TABLE_USERS + " WHERE " 
       + COLUMN_USERNAME + " = " + user.getUsername(); 

     SQLiteDatabase db = this.getWritableDatabase(); 

     Cursor cursor = db.rawQuery(selectQuery, null); 

     User findUser = new User(); 

     if (cursor.moveToFirst()) { 
      cursor.moveToFirst(); 
      findUser.setUserName(cursor.getString(1)); 
      findUser.setPassword(cursor.getString(2)); 
      cursor.close(); 
     } else { 
      findUser = null; 
     } 
     db.close(); 
     return findUser; 

    } 

}

当我尝试运行findUser(User user)方法，我得到一个SQLException没有这样的柱：Gebruikersnaam（注：Gebruikersnaam是user.getUserName（）的变量，我想这个函数返回一旦发现或为空时，它不是用户获取SQLiteException：没有这样的列

我在做什么错的，是有办法使用查询一种更好的方式

崩溃日志：？

07-25 15 ：22：42.345 32264-32264/com.example.hoofdgebruiker.winkelskortrijk E/AndroidRuntime：致命异常：主工艺：com.example.hoofdgebruiker.winkelskortrijk，PID：32264 android.database.sqlite.SQLiteException：没有这样的柱： Gebruikersnaam （码1）：在编译：SELECT * FROM用户WHERE 用户名= Gebruikersnaam 在 android.database.sqlite.SQLiteConnection.nativePrepareStatement（母语方法）在 android.database.sqlite.SQLiteConnection.acquirePreparedStatement（SQLiteConnection。 java：887） at（SQLiteProgram.java）SQLiteProgram.java（SQLiteProgram.java）（SQLiteProgram.java）（SQLiteProgram.java）（SQLiteProgram.java）（SQLiteProgram.java）（SQLiteProgram.java）（SQLiteProgram.java）：58）在android.database.sqlite.SQLiteQuery（SQLiteQuery.java:37）在 android.database.sqlite.SQLiteDirectCursorDriver.query（SQLiteDirectCursorDriver.java:44）在 android.database.sqlite.SQLiteDatabase。 rawQueryWithFactory（SQLiteDatabase.java:1316）在 android.database.sqlite.SQLiteDatabase.rawQuery（SQLiteDataba se.java:1255）在 com.example.hoofdgebruiker.winkelskortrijk.Database.MyDBHandler.findUser（MyDBHandler.java:78）在 com.example.hoofdgebruiker.winkelskortrijk.Homepage.MainActivity.register（MainActivity.java： 49） at com.example.hoofdgebruiker.winkelskortrijk.Homepage.Register $ 1.onClick（Register.java:47） at android.view.View.performClick（View.java:5198） at android.view.View $ PerformClick.run（View.java:21147）在android.os.Handler.handleCallback（Handler.java:739）在android.os.Handler.dispatchMessage（HANDL er.java:95） at android.os.Looper.loop（Looper.java:148） at android.app.ActivityThread.main（ActivityThread.java:5417） at java.lang.reflect.Method.invoke（本地方法） at com.android.internal.os.ZygoteInit $ MethodAndArgsCaller.run（ZygoteInit.java:726） at com.android.internal.os.ZygoteInit.main（ZygoteInit。Java的：616）

来源

2016-07-25 user3117628

请添加完整的崩溃日志 – USKMobility

@USKMobility完成 – user3117628

我建议你写完整的sql字符串以便在执行前检查它 – Vyacheslav

刚刚更新查询

String selectQuery = "SELECT * FROM " + TABLE_USERS + " WHERE " 
      + COLUMN_USERNAME + " = " + user.getUsername();

到

String selectQuery = "SELECT * FROM " + TABLE_USERS + " WHERE " 
      + COLUMN_USERNAME + " = '" + user.getUsername()+"'";

附上您的user.getUsername（）值单引号，因为你定义COLUMN_USERNAME为文本。如果你的字符串已经包含引号，那么它不起作用。在添加user.getUsername（）以查询从用户名中删除引号之前，请先删除以下用户名引用

user.getUsername（）。replace（“\'”，“''”）;

来源

2016-07-25 16:37:31 USKMobility

当名称包含报价时，这将炸毁。 –

@CL你是对的 – USKMobility

-1

你的问题是在这里：

String selectQuery = "SELECT * FROM " + TABLE_USERS + " WHERE " 
     + COLUMN_USERNAME + " = " + user.getUsername();

由于user.getUsername()是文本，你必须做到以下几点：

String selectQuery = "SELECT * FROM " + TABLE_USERS + " WHERE " 
     + COLUMN_USERNAME + " = '" + user.getUsername() + "'";

干杯

来源

2016-07-25 16:37:48 zozelfelfo

获取SQLiteException：没有这样的列

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