Question

我在SQLite中创建了一个名为用户的表格，其中包含3列，名称，密码和图片。< / p>

MyDatabaseHelper

public class MyDatabaseHelper extends SQLiteOpenHelper {
    public static final int DATABASE_VERSION=1;
    public static final String DATABASE_NAME="mm.db";
    public static final String TABLE_USER="User";
    public static final String KEY_NAME="Name";
    public static final String KEY_PASSWORD="Password";
    public static final String KEY_IMAGE="Image";
    public static final String ID="id";

    public void onCreate(SQLiteDatabase db) {
        db.execSQL("create table " + TABLE_USER + " ( " + ID + " INTEGER PRIMARY KEY ,Name TEXT,Password TEXT,Image BLOB )");
    }

    public void onUpgrade(SQLiteDatabase db, int oldVersion,int newVersion) {
        Log.w(MyDatabaseHelper.class.getName(), "Upgrading database from version" + oldVersion + "to" + newVersion + ",which will destroy all old data");
        db.execSQL("Drop TABLE IF EXISTS " + TABLE_USER);
        onCreate(db);
    }

    public MyDatabaseHelper(Context context)
    {
        super(context, DATABASE_NAME,null,1);
    }
}

当我尝试从SQLite检索图像并放入ImageView时，我收到错误，但是当我插入数据时，它运行正常。

MainActivity

public class MainActivity extends AppCompatActivity {
    EditText name;
    private MyDatabaseHelper mdb=null;
    private SQLiteDatabase database= null;
    ImageView imageView;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        mdb = new MyDatabaseHelper(this);
        database = mdb.getWritableDatabase();
        Button createAccount = (Button) findViewById(R.id.btnLinkToRegisterScreen);
        Bitmap picture = BitmapFactory.decodeResource(getResources(), R.mipmap.profile);
        imageView = (ImageView) findViewById(R.id.imgProfilePicture);
        imageView.setImageBitmap(getRoundedBitmap(picture));
        name = (EditText) findViewById(R.id.name);

        createAccount.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                String personName = name.getText().toString();
                String selectQuery = " SELECT " + MyDatabaseHelper.KEY_IMAGE + " FROM " + MyDatabaseHelper.TABLE_USER + " WHERE " + MyDatabaseHelper.KEY_NAME + " = " +personName;
                Cursor cursor = database.rawQuery(selectQuery, null);
                if (cursor.moveToFirst()) {
                    byte[] temp_image = cursor.getBlob(1);
                    Bitmap bitmap = BitmapFactory.decodeByteArray(temp_image, 0, temp_image.length);
                    imageView.setImageBitmap(bitmap);
                }
            }
        });
    }
}

错误

09-12 17:33:09.083  14231-14231/? E/AndroidRuntime﹕ FATAL EXCEPTION: main
    android.database.sqlite.SQLiteException: no such column: alex (code 1): , while compiling: SELECT Image FROM User WHERE Name = alex
            at android.database.sqlite.SQLiteConnection.nativePrepareStatement(Native Method)
            at android.database.sqlite.SQLiteConnection.acquirePreparedStatement(SQLiteConnection.java:888)
            at android.database.sqlite.SQLiteConnection.prepare(SQLiteConnection.java:499)
            at android.database.sqlite.SQLiteSession.prepare(SQLiteSession.java:588)
            at android.database.sqlite.SQLiteProgram.<init>(SQLiteProgram.java:58)
            at android.database.sqlite.SQLiteQuery.<init>(SQLiteQuery.java:37)
            at android.database.sqlite.SQLiteDirectCursorDriver.query(SQLiteDirectCursorDriver.java:44)
            at android.database.sqlite.SQLiteDatabase.rawQueryWithFactory(SQLiteDatabase.java:1314)
            at android.database.sqlite.SQLiteDatabase.rawQuery(SQLiteDatabase.java:1253)
            at com.example.seng.healthyapp.MainActivity$1.onClick(MainActivity.java:76)
            at android.view.View.performClick(View.java:4230)

alex实际上是我键入的userName，但它没有说明这样的列。

Answer 1

将您的字符串用单引号括起来，例如

SELECT Image FROM User WHERE Name = 'alex'

SQLite

1 个答案: