Question

我正在尝试在android中执行基本登录表单。这是代码的修剪版本。基本上，如果用户输入无效的登录详细信息，则应触发警报。但是，当我点击登录按钮时，我收到一个错误。下面发布的是我的代码。

AWSDynamoDBPutItemInput *request = [[AWSDynamoDBPutItemInput alloc] init];

发布的是我的数据库助手类

Alert_Dialouge alert = new Alert_Dialouge();
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    setContentView(R.layout.activity_main_activity);
    context = this;
    final DataBaseHelper hl =  new DataBaseHelper(context);
    AlertDialog alertDialog = new AlertDialog.Builder(
            Main_activity.this).create();
    Button b1 = (Button) findViewById(R.id.login);
    v1 = (TextView) findViewById(R.id.TEXT_STATUS_ID);
    b1.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {

            c = c + 1;
            EditText user = (EditText) findViewById(R.id.username);
            EditText pwd = (EditText) findViewById(R.id.password);
            String username = user.getText().toString();
            String password = pwd.getText().toString();

            db = hl.getReadableDatabase();
            db.beginTransaction();
            String select = "SELECT * from  " + " " + hl.TABLE_NAME+ " "+ 
            "where Username = admin and Password = admin;" ;
            Cursor c = db.rawQuery(select, null);
            if ((username.equals("") && password.equals("")) ||  
            password.equals("") || username.equals("")||c.getCount()==0) {

                alert.showAlertDialog(Main_activity.this, "Please enter your credentials", false);
                Intent intent = new Intent(getApplicationContext(), Main_activity.class);
                startActivity(intent);
                finish();

            } else if(c.getCount()>0) {


                      c.moveToFirst();
                System.out.println("hiiii");

            }
            c.close();
        }
       });
        }
       }

发布的是stackTrace：

      public class DataBaseHelper extends SQLiteOpenHelper {
      public static final String DATABASE_NAME ="db1";
     public static  final int DATABASE_VERSION=8;
     public static final String COLUMN_ID = "_id";
     public static final String TABLE_NAME ="malware_users1";
public static final String CREATE_TABLE = "CREATE TABLE IF NOT EXISTS " +"  
"+ TABLE_NAME +"( _id INTEGER PRIMARY KEY AUTOINCREMENT,Username 
 VARCHAR(255),Password VARCHAR(255))";


     public static final String DELETE_TABLE= "DROP TABLE IF EXISTS " +  
     TABLE_NAME;


    public DataBaseHelper(Context context) {
    super(context,DATABASE_NAME,null,DATABASE_VERSION);
      }
     @Override
      public void onCreate(SQLiteDatabase db) {
        db.execSQL(CREATE_TABLE);
      }

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

    db.execSQL(DELETE_TABLE);
    oncreate(db);
      }


  public void insertData(String Username,String Password)
    {
    SQLiteDatabase db = this.getWritableDatabase();
    db.beginTransaction();
    ContentValues values;

    try
    {
        values = new ContentValues();
        values.put("Username",Username);
        values.put("Password",Password);
        long i =db.insert(TABLE_NAME,null,values);
        Log.i("Insert",i+"");
        db.setTransactionSuccessful();
     }
    catch (SQLiteException e)
    {
        e.printStackTrace();
    }
    finally {
        db.endTransaction();
        db.close();
                    }
                   }
                  }

Answer 1

请检查我更改的查询

String select = "SELECT * from  " + " " + hl.TABLE_NAME+ " "+ 
            "where Username = 'admin' and Password = 'admin'" ;

请检查一下，让我知道它解决了这个问题吗？

Answer 2

更好的方法是使用selectionArgs参数

 String select = "SELECT * from  " + " " + hl.TABLE_NAME+ " "+ 
                                      "where Username = ? and Password = ? "; 
 Cursor c = db.rawQuery(select, new String[] {"admin", "admin});

Here您可以找到文档

Answer 3

你的b1 clicklistener中的

问题出在你的选择查询

上

更改此

   String select = "SELECT * from  " + " " + hl.TABLE_NAME+ " "+ 
        "where Username = admin and Password = admin;" ;

到

  String username="admin";
  String password="admin";
   String select = "SELECT * from " + hl.TABLE_NAME+
            " where Username ='"+username+"' and Password ='"+password+"'" ;

SQLiteLog：没有这样的列

3 个答案: