Question

我正在尝试从“users”表中获取所有用户的列表，我收到以下错误：

org.hibernate.hql.internal.ast.QuerySyntaxException: users is not mapped [from users]
org.hibernate.hql.internal.ast.util.SessionFactoryHelper.requireClassPersister(SessionFactoryHelper.java:180)
org.hibernate.hql.internal.ast.tree.FromElementFactory.addFromElement(FromElementFactory.java:110)
org.hibernate.hql.internal.ast.tree.FromClause.addFromElement(FromClause.java:93)

这是我写的添加/获取用户的代码：

public List<User> getUsers() {
    Session session = HibernateUtil.getSessionFactory().getCurrentSession();
    session.beginTransaction();
    List<User> result = (List<User>) session.createQuery("from users").list();
    session.getTransaction().commit();
    return result;
}

public void addUser(User user) {
    Session session = HibernateUtil.getSessionFactory().getCurrentSession();
    session.beginTransaction();
    session.save(user);
    session.getTransaction().commit();
}

public void addUser(List<User> users) {
    Session session = HibernateUtil.getSessionFactory().getCurrentSession();
    session.beginTransaction();
    for (User user : users) {
        session.save(user);
    }
    session.getTransaction().commit();
}

添加用户可以正常工作，但是当我使用getUsers函数时，我会收到这些错误。

这是我的hibernate配置文件：

<hibernate-configuration>
<session-factory>
    <property name="connection.url">jdbc:mysql://localhost:3306/test</property>
    <property name="connection.username">root</property>
    <property name="connection.password">root</property>
    <property name="connection.driver_class">com.mysql.jdbc.Driver</property>
    <property name="hibernate.default_schema">test</property>
    <property name="dialect">org.hibernate.dialect.MySQL5Dialect</property>

    <property name="show_sql">true</property>

    <property name="format_sql">true</property>
    <property name="hbm2ddl.auto">create-drop</property>

    <!-- JDBC connection pool (use the built-in) -->
    <property name="connection.pool_size">1</property>
    <property name="current_session_context_class">thread</property>

    <!-- Mapping files will go here.... -->

    <mapping class="model.Company" />
    <mapping class="model.Conference" />
    <mapping class="model.ConferencesParticipants" />
    <mapping class="model.ConferenceParticipantStatus" />
    <mapping class="model.ConferencesUsers" />
    <mapping class="model.Location" />
    <mapping class="model.User" />

</session-factory>

这是我的用户类：

@Entity
@Table( name = "Users" )
public class User implements Serializable{

    private long userID;
    private int pasportID;
    private Company company; 
    private String name;
    private String email;
    private String phone1;
    private String phone2;
    private String password; //may be null/empty , will be kept hashed
    private boolean isAdmin;
    private Date lastLogin;

    User() {} //not public on purpose!

    public User(int countryID, Company company, String name, String email,
            String phone1, String phone2, String password, boolean isAdmin) {
        this.pasportID = countryID;
        this.company = company;
        this.name = name;
        this.email = email;
        this.phone1 = phone1;
        this.phone2 = phone2;
        this.password = password;
        this.isAdmin = isAdmin;
    }

    @Id
    @GeneratedValue(generator="increment")
    @GenericGenerator(name="increment", strategy = "increment")
    public long getUserID() {
        return userID;
    }
    public void setUserID(long userID) {
        this.userID = userID;
    }
    ...    
}

知道我为什么会收到这个错误吗？

Answer 1

在HQL中，您应该使用映射的@Entity的 java类名和属性名，而不是实际的表名和列名，所以HQL应该是：

List<User> result = (List<User>) session.createQuery("from User").list();

Answer 2

例如：您的bean类名称是 UserDetails

Query query = entityManager. createQuery("Select UserName from **UserDetails** ");

您不会在Db上提供您的表名。 你给出了bean的类名。

Answer 3

只是为了分享我的发现。即使查询的目标是正确的类名，我仍然会遇到相同的错误。后来我意识到我从错误的包中导入了Entity类。

我从以下位置更改导入行后问题得以解决：

import org.hibernate.annotations.Entity;

到

import javax.persistence.Entity;

Answer 4

添加@TABLE(name = "TABLE_NAME")注释并修复。检查您的注释和hibernate.cfg.xml文件。这是有效的示例实体文件：

import javax.persistence.*;

@Entity
@Table(name = "VENDOR")
public class Vendor {

    //~ --- [INSTANCE FIELDS] ------------------------------------------------------------------------------------------
    private int    id;
    private String name;

    //~ --- [METHODS] --------------------------------------------------------------------------------------------------
    @Override
    public boolean equals(final Object o) {    
        if (this == o) {
            return true;
        }

        if (o == null || getClass() != o.getClass()) {
            return false;
        }

        final Vendor vendor = (Vendor) o;

        if (id != vendor.id) {
            return false;
        }

        if (name != null ? !name.equals(vendor.name) : vendor.name != null) {
            return false;
        }

        return true;
    }

    //~ ----------------------------------------------------------------------------------------------------------------
    @Column(name = "ID")
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Id
    public int getId() {
        return id;
    }

    @Basic
    @Column(name = "NAME")
    public String getName() {

        return name;
    }

    public void setId(final int id) {
        this.id = id;
    }

    public void setName(final String name) {    
        this.name = name;
    }

    @Override
    public int hashCode() {
        int result = id;
        result = 31 * result + (name != null ? name.hashCode() : 0);
        return result;
    }
}

Answer 5

某些基于Linux的MySQL安装需要区分大小写。解决方法是应用nativeQuery。

@Query(value = 'select ID, CLUMN2, CLUMN3 FROM VENDOR c where c.ID = :ID', nativeQuery = true)

Answer 6

org.hibernate.hql.internal.ast.QuerySyntaxException: users is not mapped [from users]

这表明hibernate不知道User实体是“用户”。

@javax.persistence.Entity
@javax.persistence.Table(name = "Users")
public class User {

@Table注释将表名称设置为“用户”，但实体名称仍在HQL中称为“用户”。

要更改两者，您应该设置实体的名称：

// this sets the name of the table and the name of the entity
@javax.persistence.Entity(name = "Users")
public class User implements Serializable{

有关详情，请参阅我的回答：Hibernate table not mapped error

Answer 7

还要确保在hibernate bean配置中设置了以下属性：

<property name="packagesToScan" value="yourpackage" />

这告诉spring和hibernate在哪里找到注释为实体的域类。

Answer 8

您可能忘记将创建的实体的映射添加到hibernate.cfg.xml，同样的错误。

Answer 9

还要检查是否使用以下方法添加了带注释的类：

new Configuration().configure("configuration file path").addAnnotatedClass(User.class)

使用Hibernate在数据库中添加新表时，总是浪费我的时间。

Answer 10

我也导入了错误的实体import org.hibernate.annotations.Entity; 它应该是导入javax.persistence.Entity;

Answer 11

当我用hibernate-core-5.2.12替换旧的hibernate-core库时，我遇到了这个问题。但是我的所有配置都没问题。我通过这种方式创建sessionfactory来解决这个问题：

private static SessionFactory buildsSessionFactory() {
    try {
        if (sessionFactory == null) {
            StandardServiceRegistry standardRegistry = new StandardServiceRegistryBuilder()
                    .configure("/hibernate.cfg.xml").build();
            Metadata metaData = new MetadataSources(standardRegistry)
                    .getMetadataBuilder().build();
            sessionFactory = metaData.getSessionFactoryBuilder().build();
        }
        return sessionFactory;
    } catch (Throwable th) {

        System.err.println("Enitial SessionFactory creation failed" + th);

        throw new ExceptionInInitializerError(th);

    }
}

希望它有助于某人

Answer 12

使用Spring休眠时，我遇到相同的错误。.我在createQuery语句中使用小写字母的“ user”，而我的班级是User.。因此在查询中将其更改为User，问题是解决。

之前查询：

Query query= em.createQuery("select u from user u where u.username=:usr AND u.password=:pass",User.class);

查询之后：

Query query= em.createQuery("select u from User u where u.username=:usr AND u.password=:pass",User.class);

Answer 13

我推荐这种模式：

@Entity(name = User.PERSISTANCE_NAME)
@Table(name = User.PERSISTANCE_NAME )
public class User {    
    static final String PERSISTANCE_NAME = "USER";

    // Column definitions here

}

Answer 14

在查询中，您必须使用类名（用户）而不是表名（用户）所以你的查询是 “来自用户”

Answer 15

您必须在选择查询中输入与实体或类相同的名称（区分大小写）。即从className / Entity Name用户中选择用户；

Answer 16

使用org.hibernate.hql.internal.ast.QuerySyntaxException: users is not mapped [from users]，您正在尝试从users表中进行选择。但是您正在用@Table( name = "Users" )注释您的班级。因此，请使用users或Users。

Answer 17

如果使用xml配置，则applicationContext.xml文件中将需要以下内容：

<bean id="sessionFactory" class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean" lazy-init="default" autowire="default" dependency-check="default">
<property name="dataSource">
  <ref bean="dataSource" />
</property>
<property name="annotatedClasses">
  <list>
    <value>org.browsexml.timesheetjob.model.PositionCode</value>
  </list>

Answer 18

在使用Quarkus微服务框架时，我也遇到了这个问题：

public class SomeResource {

  @GET
  @RolesAllowed({"basic"})
  public Response doSomething(@Context SecurityContext context) {
    // ...
  }
}

// this will generate an QuerySyntax exception, as the authorization module
// will ignore the Entity annotation and use the class name instead.
@Entity(name = "users")
@UserDefinition
public class User {
  // ...
}

// do this instead
@Entity
@Table(name = "users")
@UserDefinition
public class User {
  // ...
}

Answer 19

在Spring项目中：我输入了错误的CODEPIPELINE并收到此错误。现在运行良好。

Hibernate错误 - QuerySyntaxException：用户未映射[来自用户]

19 个答案: