在Android Studio中连接外部SQLite数据库，需要一些指导

Question

这是我的数据库设计。我有2个表“ Donation_Details ”和“ 信息”。让我澄清一下，现在我不能把这两个表叫出来了，这是什么问题？我被困在这里将近1天...我试图自己找到答案，但根本没有答案...

这是我的DatabaseOpenHelper

public class DatabaseOpenHelper extends SQLiteAssetHelper {

    private static final String DATABASE_NAME = "eBossCharity.db";
    private static final int DATABASE_VERSION = 1;

    public DatabaseOpenHelper(Context context) {
        super(context, DATABASE_NAME, null, DATABASE_VERSION);
    }

这是我的数据库访问权限

public class DatabaseAccess {
    private SQLiteOpenHelper openHelper;
    private SQLiteDatabase db;
    private static DatabaseAccess instance;
    Cursor c = null;

    private DatabaseAccess(Context context) {
        this.openHelper = new DatabaseOpenHelper(context);
    }

    public static DatabaseAccess getInstance(Context context) {
        if (instance == null) {
            instance = new DatabaseAccess(context);
        }
        return instance;
    }

    public void open() {
        this.db = openHelper.getWritableDatabase();
    }

    public void close() {
        if (db != null) {
            this.db.close();
        }
    }

    public String getAddress(String name){
        c=db.rawQuery("Select * from Information ",new String[]{});
        StringBuffer buffer =new StringBuffer();
        while (c.moveToNext()){
            String address = c.getString(0);
            String address1 = c.getString(1);
            buffer.append(" "+ address +" "+ address1);
        }
        return buffer.toString();
    }
}

这是我的Java页面

public void onClick(View v) {
                DatabaseAccess databaseAccess = DatabaseAccess.getInstance(getApplicationContext());
                databaseAccess.open();

                try {
                    String n = summary_showAll.getText().toString();
                    String address = databaseAccess.getAddress(n);

                    summary_showAll.setText(address);
                    databaseAccess.close();

                } catch (Exception e) {
                    e.printStackTrace();
                }


            }

我是否键入错误导致此错误？

Answer 1

我也有同样的情况。您所需要做的就是卸载和重新安装您的应用。它为我工作。
希望有帮助。