java - Hql 错误 : Class is not mapped, 无法解析符号

Question

我正在学习 Hibernate，并且在 hql 方面遇到了困难。我希望我的函数检查数据库中是否存在用户名。

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;
    }

上面的函数位于我的 UserControl 类中。这是我在 IntelliJ 中的项目布局:

在我的 UserControl 类中，我已经导入了 User 类，例如 importentity.User，但我仍然无法在 HQL 中使用裸露的 User 类名来代替 entity.User 不会出现以下错误。

java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: User is not mapped [select 1 from User u where u.userName = :userName]

我搜索类似问题时发现了this answer以及页面上的其他内容表明我的实体类的命名存在一些错误，尽管我基于答案的实验没有成功。如果我屈服并像上面那样使用 entity.User ，那么我会收到此错误:

java.lang.IllegalArgumentException: Could not locate named parameter [userName], expecting one of []

这是我的实体类:

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;
    }

    @Column(name = "weight")
    public Double getWeight(){ return weight; }
    public void setWeight(Double weight){ this.weight = weight; }
}

还有我的 hibernate.cfg.xml

<?xml version='1.0' encoding='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
    "-//Hibernate/Hibernate Configuration DTD//EN"
    "http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
  <session-factory>
    <property name="connection.url">
      jdbc:mysql://localhost:3306/login_register_view?createDatabaseIfNotExist=true</property>
    <property name="connection.driver_class">
      com.mysql.cj.jdbc.Driver</property>
    <property name="connection.username">root</property>
    <property name="connection.password">pass</property>
    <property name="hibernate.dialect">
      org.hibernate.dialect.MySQL8Dialect</property>
    <property name="show_sql">true</property>
    <property name="format_sql">true</property>
    <property name="hbm2ddl.auto">update</property>
  </session-factory>
</hibernate-configuration>

我的 web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd"
         version="4.0">

    <servlet>
        <servlet-name>UserControl</servlet-name>
        <servlet-class>control.UserControl</servlet-class>
    </servlet>

    <servlet-mapping>
        <servlet-name>UserControl</servlet-name>
        <url-pattern>/user-control</url-pattern>
    </servlet-mapping>

</web-app>

我做错了什么？

Answer 1

我发现发生这种情况的原因是因为我没有在 hibernate.cfg.xml 中映射该实体。里面<session-factory> </session-factory>我需要写:

<mapping class="entity.User"/>