Android sqlite 登录错误 activity

Question

当用户登录时，应用程序似乎崩溃了，然后又工作了。 LogCat 上出现错误：

Caused by: android.database.sqlite.SQLiteException: no such column: cliff (code 1): , while compiling: SELECT DISTINCT _id, name, password FROM Users WHERE name=cliff

为什么它会抛出一个错误但又能正常工作？谁能看到我的问题。下面是我的代码...

 public class DBAdapter {
static final String KEY_ROWID = "_id";
static final String KEY_USERNAME = "name";
static final String KEY_PASSWORD = "password";
static final String KEY_CARDNUM = "cardnum";
static final String KEY_CARDNAME = "cardname";
static final String TAG = "DBAdapter";

static final String DATABASE_NAME = "CPDB";
static final String DATABASE_TABLE = "Users";
static final String DATABASE_TABLEC = "Cards";
static final int DATABASE_VERSION = 1;

static final String DATABASE_CREATE =
        "create table Users (_id integer primary key autoincrement, " + "name text not null, password text not null, cardnum integer not null, cardname text not null);";


final Context context;

DatabaseHelper DBHelper;
SQLiteDatabase db;

public DBAdapter (Context ctx){
    this.context = ctx;
    DBHelper = new DatabaseHelper(context);
}

private static class DatabaseHelper extends SQLiteOpenHelper
{
    DatabaseHelper(Context context){
        super (context, DATABASE_NAME, null, DATABASE_VERSION);
    }

    @Override
    public void onCreate(SQLiteDatabase db){
        try{
            db.execSQL(DATABASE_CREATE);
        } catch (SQLException e) {
            e.printStackTrace();
        }
    }

    @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
        Log.w(TAG, "Upgrading database from version" + oldVersion + "to" + newVersion + "which will destroy all old data");
        db.execSQL("DROP TABLE IF EXISTS Users");
        onCreate (db);

    }
}

//--opens the DB--
public DBAdapter open() throws SQLException
{
    db= DBHelper.getWritableDatabase();
    return this;
}

//--close the DB--
public void close()
{
    DBHelper.close();
 } 

//--retrieves chosen user--
  public Cursor checkUser(String username) throws SQLException{
    Cursor mCursor =
            db.query(true, DATABASE_TABLE, new String[] {KEY_ROWID,       KEY_USERNAME, KEY_PASSWORD}, KEY_USERNAME + "=" + username, null, null, null, null,null);
    if (mCursor != null) {
        mCursor.moveToFirst();
    }
    return mCursor;
}

}

然后进行登录javaclass......

 public class login extends Activity {
Button btn1,btn2;

@Override
public void onCreate (Bundle savedInstanceState) {
    /* Called when the activity is first created*/
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);
    btn1 = (Button)findViewById(R.id.bttn_signup);
    btn2 = (Button)findViewById(R.id.bttn_signin);
}

 public void onClick(View view){

    if(view.getId() == R.id.bttn_signup) {
        startActivity(new Intent("com.cardpocket.cardpocket.signup"));
    }

    if(view.getId() == R.id.bttn_signin) {

     EditText email = (EditText) findViewById(R.id.tf_email);
        EditText pass = (EditText) findViewById(R.id.tf_password);

        String username = email.getText().toString();
        String password = pass.getText().toString();


        DBAdapter db = new DBAdapter(this);

        db.open();
        Cursor c = db.checkUser(username);
        //set someUsers as a list
        ArrayList<String> theUser = new ArrayList<>();


        if (c.moveToFirst()) {
            do {
                String aUser = c.getString(c.getColumnIndex("password"));
                theUser.add(aUser);


            } while (c.moveToNext());
        }

      if (theUser.contains(password)) {

          startActivity(new Intent("com.cardpocket.cardpocket.mainactivity"));

          /*
       //set the value and pass to the mainactivity
          Intent i = new Intent(this, mainactivity.class);
          i.putExtra("username", username);
          startActivity(i);}*/

      } else {
          //do something else
        Toast.makeText(this, "No contact found", Toast.LENGTH_LONG).show();
      }
    }
}

提前谢谢你。 :)

Answer 1

您容易受到 SQL injection attacks 的影响，忘记引用您的价值观：

SELECT DISTINCT _id, name, password FROM Users WHERE name='cliff'
                                                          ^-----^---

没有引号，cliff 被视为不存在的 field/table 名称。带引号的 'cliff' 是字符串文字。

Answer 2

错误很明显："cliff" 列不存在...

您必须在 SQL 请求

中的 cliff 周围加上引号

db.query(true, DATABASE_TABLE, new String[] {KEY_ROWID,       KEY_USERNAME, KEY_PASSWORD}, KEY_USERNAME + "= '" + username + "'", null, null, null, null,null);

此外，我建议您使用准备好的 SQL 查询，这样：

String selection = "KEY_USERNAME = ?";
String[] selectionArgs = { username };
String[] columns = {KEY_ROWID, KEY_USERNAME, KEY_PASSWORD};
Cursor cursor = db.query(true, DATABASE_TABLE, columns, selection, selectionArgs , null, null, null,null);

它会为您添加引号，确保您的请求更安全。

Link : Official documentation of query method

Answer 3

您必须使用（引号）'cliff' 而不是悬崖。如果没有引号，它将被 sqlite 识别为列，并且可能会给出错误 no such column (cliff)。

Android sqlite 登录错误 activity

Android sqlite error for login activity

java

sqlite

android

android-sqlite