Interroger base de données avec hibernate

package hacktrack.login;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Table;

@Entity(name = "User")
@Table(name = "T_USERS")
public class User {
	@Column(name = "USER_NAME")
	private String username;

	@Column(name = "PASSWORD")
	private String password;

	public String getUsername() {
		return username;
	}

	public void setUsername(String username) {
		this.username = username;
	}

	public String getPassword() {
		return password;
	}

	public void setPassword(String password) {
		this.password = password;
	}
}

package hacktrack.login;

import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Query;

public class Login {

	@Autowired//Annotation de Spring permettant d'injecter dynamiquement le EntityManagerFactory (à configurer dans le contexte de Spring)
	EntityManagerFactory emf;
	
	public void login(String username, String password) throws Exception{
		EntityManager em = emf.createEntityManager();
		Query q = em.createQuery("FROM User u WHERE u.username='"+username+"'");//Requête JPQL permettant de récupérer le user portant ce nom
		User user = (User)q.getSingleResult();
		if(!checkPassword(user, password))
			throw new Exception("Login failed");
		//Effectue ici toutes les actions nécessaires lorssque le login a réussi
		//...
	}
	
	private boolean checkPassword(User user, String password){
		return user.getPassword().equals(password);
	}
}