我正在學習 Hibernate，并且在使用 hql 時遇到了困難。我希望我的函數檢查數據庫中是否存在用戶名。private boolean userExists(Session session, String userName) {        String hql = "select 1 from entity.User u where u.userName = :userName";        Query query = session.createQuery(hql);        query.setParameter("userName", userName);        return query.uniqueResult() != null;    }上面的函數位于我的 UserControl 類中。這是我在 IntelliJ 中的項目布局：在我的 UserControl 類中，我已經導入了我的 User 類import entity.User，但我仍然不能User在 HQL 中使用裸類名而不是entity.User沒有得到以下錯誤。java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: User is not mapped [select 1 from User u where u.userName = :userName]我對類似問題的搜索讓我找到了這個答案和頁面上的其他人，這表明我的實體類的命名存在一些錯誤，盡管我基于答案的實驗沒有奏效。entity.User如果我像上面那樣屈服并使用，那么我會收到此錯誤：java.lang.IllegalArgumentException: Could not locate named parameter [userName], expecting one of []這是我的實體類：package entity;import javax.persistence.Column;import javax.persistence.Entity;import javax.persistence.Id;import javax.persistence.Table;@Entity@Table(name = "user_association", schema = "login_register_view")public class User {    private int id;    private String userName;    private String password;    private String color;    private Integer pocketCount;    private Double weight;    @Id    @Column(name= "id", nullable = false)    public int getId(){ return id; }    public void setId(int id){ this.id = id;}    @Column(name = "user_name")    public String getUserName(){ return userName; }    public void setUserName(String userName){ this.userName = userName; }    @Column(name = "password")    public String getPassword(){ return password; }    public void setPassword(String password){ this.password = password; }    @Column(name = "color")    public String getColor(){ return color; }    public void setColor(String color){ this.color = color; }    @Column(name = "pocket_count")    public Integer getPocketCount(){ return pocketCount; }    public void setPocketCount(Integer pocketCount){        this.pocketCount = pocketCount;    }}