Hql错误：未映射类，无法解析符号

Question

我正在学习Hibernate，并且在使用hql时遇到困难。我希望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;
    }

以上功能位于我的UserControl类中。这是我在IntelliJ中的项目布局：

在UserControl类中，我已经导入了import entity.User之类的User类，但是仍然无法在HQL中使用裸露的User类名而不是entity.User，而不会出现以下错误

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

我对类似问题的搜索使我指向this answer，页面上的其他问题则表明我的Entity类的命名存在一些错误，尽管我根据答案进行的实验没有用。如果我像上面那样屈服并使用entity.User，则会收到此错误：

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

这是我的Entity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>

我怎么了？

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。在<session-factory> </session-factory>里面，我需要写：

<mapping class="entity.User"/>