Question

我已经实现了一种针对SQLite DB验证用户的方法，但是单击登录按钮时，程序始终引发错误，错误代码如下

2019-04-15 21：38：27.455 4601-4601 / com.example.mymedicare E / AndroidRuntime：致命异常：main 流程：com.example.mymedicare，PID：4601 android.database.sqlite.SQLiteException：无此类列：Brad（代码1）：，而在编译时：SELECT * FROM users WHERE Brad =？和 Brad1994 =？在android.database.sqlite.SQLiteConnection.nativePrepareStatement（Native 方法）在android.database.sqlite.SQLiteConnection.acquirePreparedStatement（SQLiteConnection.java:889）在android.database.sqlite.SQLiteConnection.prepare（SQLiteConnection.java:500）在android.database.sqlite.SQLiteSession.prepare（SQLiteSession.java:588）在android.database.sqlite.SQLiteProgram。（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:1318）在android.database.sqlite.SQLiteDatabase.rawQuery（SQLiteDatabase.java:1257）在com.example.mymedicare.MyDBHelper.checkLogin（MyDBHelper.java:96）

我已经尝试读取错误日志，但是找不到任何严重错误，我的方法将显示在下面

  // This is my check login method in mydbhelper   

   public boolean checkLogin(String username, String password) {
    SQLiteDatabase db = this.getWritableDatabase();

    String s;
    Cursor c = db.rawQuery("SELECT * FROM users WHERE " + username + " =? AND " + password + " =?", null);

    if (c.getCount() <= 0) {
        c.close();
        db.close();
        return false;
    } else {
        c.close();
        db.close();
        return true;
    }
}


   //This is LoggingIn activity where I have tried to implement the above 
   method 

   public class LoggingIn extends AppCompatActivity {

//defines page objects
private Button SignIn;
private EditText Usernames;
private EditText Passwords;
private MyDBHelper db;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_logging_in);

    //Finds objects by ID
     //edit text to string
    Usernames = (EditText) findViewById(R.id.Usernames);

    Passwords = (EditText) findViewById(R.id.Passwords);


    Button SignIn = (Button) findViewById(R.id.SignIn);
    db = new MyDBHelper(this);

    SignIn.setOnClickListener(new Button.OnClickListener() {
        @Override
        public void onClick(View v) {
            Boolean validUser = 
     db.checkLogin(Usernames.getText().toString(), 
     Passwords.getText().toString());

            if (validUser == true) {
                Intent i = new Intent(getApplicationContext(), 
                HomePage.class);
                startActivity(i);
            } else {
                Toast.makeText(getApplicationContext(), "Invalid login", 
                Toast.LENGTH_SHORT).show();
            }

        }
    });
}
}

Answer 1

您正在传递参数username和password作为列名。
正确的方法是：

Cursor c = db.rawQuery(
    "SELECT * FROM users WHERE usenamecolumn = ? AND passwordcolumn = ?", 
     new String[] {username, password});

将usenamecolumn和passwordcolumn更改为表的列名，以获取用户名和密码。

Answer 2

您可能想要更接近此的东西

const arr = [
   {  
     "name":"BMW",
     "price":"55 000",
     "country":"Germany",
     "certificate":"yes"
  },
  {  
    "name":"Mitsubishi",
    "price":"93 000", 
    "constructor":"Bar John",
    "door":"3",
    "country":"Japan",
  },
  {  
    "name":"TOYOTA", 
    "price":"48 000", 
    "max_people":"7",
    "country":"Japan",
    "certificate":"yes"
  },
  {  
    "name":"Volkswagen",
    "price":"36 000", 
    "constructor":"Pier Sun",
    "country":"Germany",
    "certificate":"no"
  },
 ];
function pushArr (arr, item){
    arr.push(item); 
}

let Cars = pushArr(arr,{  
  "name":"Audi",
  "price":"89 000", 
  "constructor":"Hubert Trumpius",
  "country":"Germany",
  "certificate":"yes"
});

console.log(arr);

按照上面的编写方式，您尝试使用搜索条件（值）作为列名（键）来执行查询。

例如用简单的英语来说，原始语句的第一部分显示“在Brad列中找到内容”，这就是为什么您的异常会弹出“ SQLiteException：无此类列：Brad”的原因

您实际上想说的是“在列用户名中找到Brad”，这就是我在上面添加的修订后的语句应该为您所做的（未经测试，但在正确的方法上可以让您前进）。

我发现了另外一个StackOverflow帖子，它也可能有助于创建准备好的语句-> How do I use prepared statements in SQlite in Android?

希望这会有所帮助。

是什么导致以下Java代码中的错误

2 个答案: