谁能告诉我应该如何更改查询以便将数据库中的值与字符串进行比较

Question

所以我试图从我的数据库中找到一个用户，他的名称和密码字段与作为参数给出的 n 和 p 字符串相同。 这是我的 findBy function：

public User findBy(String n, String p){
            try(Session session = sessionFactory.openSession()) {
                Transaction tx = null;
                try {
                    tx = session.beginTransaction();
                    User uss = new User(n,p,TipUser.ANGAJAT);
                    User us =
                            session.createQuery("from User as u where u.name = " + uss.getName() + " and u.passwd = " + uss.getPasswd(), User.class)
                                    .uniqueResult();
                    System.out.println(us + " was found!");
                    tx.commit();
    //                return us;
                } catch (RuntimeException ex) {
                    if (tx != null)
                        tx.rollback();
                }
            }
            return null;
        }

我在主要的 function 中这样称呼它：

public static void main(String[] args) {
        try {
            initialize();

            UserRepoORM test = new UserRepoORM();
            test.findBy("Luca","luca");
        }catch (Exception e){
            System.err.println("Exception "+e);
            e.printStackTrace();
        }finally {
            close();
        }
    }

问题是我遇到的是这个异常：

22 Apr 2021 19:16:25,995 DEBUG org.hibernate.engine.jdbc.spi.SqlExceptionHelper 126 logExceptions - could not prepare statement [select user0_.id as id1_1_, user0_.Name as name2_1_, user0_.Passwd as passwd3_1_, user0_.Tip as tip4_1_, user0_.LogIn as login5_1_ from User user0_ where user0_.Name=Luca and user0_.Passwd=luca]
org.sqlite.SQLiteException: [SQLITE_ERROR] SQL error or missing database (no such column: Luca)

Answer 1

当查询具有某些参数时，您必须使用Query#setParameter而不是字符串连接来设置它们

Query<User> query = session.createQuery("from User as u where u.name=:name and u.passwd=:passwd", User.class);
query.setParameter("name", uss.getName());
query.setParameter("passwd", uss.getPasswd());
User user = query.uniqueResult();