繁体   English   中英

休眠:条件和HQL的映射行为之间的区别

[英]Hibernate: Difference between mapping behaviour of Criteria and HQL

Hibernate和HQL的Criteria API在加载类映射方面的行为有所不同。 如果我从XML加载映射,则两种方法都可以加载Entity类。 但是,如果我使用注释,则只有Criteria api能够加载映射。 请帮助我了解原因。

我正在使用Hibernate版本:

<dependency>
    <groupId>org.hibernate</groupId>
    <artifactId>hibernate-core</artifactId>
    <version>4.3.11.Final</version>
</dependency>

休眠配置

<hibernate-configuration>
    <session-factory>
        <property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>
        <property name="hibernate.dialect">org.hibernate.dialect.MySQLDialect</property>
        <property name="hibernate.connection.url">jdbc:mysql://localhost:3306/demo</property>
        <property name="hibernate.connection.username">user</property>
        <property name="hibernate.connection.password">password</property>
        <property name="hibernate.connection.pool_size">1</property>
        <property name="hibernate.current_session_context_class">thread</property>
        <property name="hibernate.show_sql">true</property>
        <property name="hibernate.archive.autodetection">class, hbm</property>

        <!--<mapping  class="com.mk.hibernate.User" resource="User.hbm.xml"/>-->
        <mapping class="com.mk.hibernate.User"/>
    </session-factory>
</hibernate-configuration>

休眠映射文件:

<?xml version="1.0" encoding="UTF-8"?>

<hibernate-mapping
                   xmlns="http://www.hibernate.org/xsd/hibernate-mapping"
                   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
                   xsi:schemaLocation="http://www.hibernate.org/xsd/hibernate-mapping
        http://www.hibernate.org/xsd/hibernate-mapping/hibernate-mapping-4.0.xsd">

    <class name="com.mk.hibernate.User" table="USERS">
        <id name="username" column="USERNAME" type="java.lang.String" >
            <generator class="assigned"></generator>
        </id>
        <property name="firstname" column="FIRSTNAME" type="java.lang.String"/>
        <property name="lastname" column="LASTNAME" type="java.lang.String"/>
        <property name="doj" column="DOJ" type="java.util.Date"/>
    </class>

</hibernate-mapping>

我有一个实体

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import java.util.Date;
import java.util.Objects;

@Entity(name = "USERS")
public class User implements java.io.Serializable {
    @Id
    @Column(name = "USERNAME")
     private String username;
    @Column(name = "FIRSTNAME")
     private String firstname;
    @Column(name = "LASTNAME")
     private String lastname;
    @Column(name = "DOJ")
     private Date doj;
...
...

我有两种方法来获取该实体的数据

public List<User> getAllUsersByCriteria() {
        Session session = sessionFactory.openSession();
        Criteria cr = session.createCriteria(User.class);
        List<User> users = cr.list();
        System.out.println(users.size());
        session.close();
        return users;
    }

 public List<User> getAllUsers() {
        Session session = sessionFactory.openSession();
        List<User> users = session.createQuery(" from User").list();
        session.close();
        return users;
    }

当我执行以下代码

        System.out.println("====By Criteria===");
        controller.getAllUsersByCriteria().forEach(u -> {
            System.out.println(u.getFirstname());
        });
        System.out.println("====By Query===");
        controller.getAllUsers().forEach(u -> {
            System.out.println(u.getFirstname());
        });

我收到以下异常

====By Criteria===
Hibernate: select this_.USERNAME as USERNAME1_0_0_, this_.DOJ as DOJ2_0_0_, this_.FIRSTNAME as FIRSTNAM3_0_0_, this_.LASTNAME as LASTNAME4_0_0_ from USERS this_
2
User1
User2
====By Query===
problem creating session factory!
org.hibernate.hql.internal.ast.QuerySyntaxException: User is not mapped [ from User]
    at org.hibernate.hql.internal.ast.QuerySyntaxException.generateQueryException(QuerySyntaxException.java:96)
    at org.hibernate.QueryException.wrapWithQueryString(QueryException.java:120)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:234)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:158)
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:131)
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:93)
    at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:167)
    at org.hibernate.internal.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:301)
    at org.hibernate.internal.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:236)
    at org.hibernate.internal.SessionImpl.createQuery(SessionImpl.java:1836)
    at com.mk.hibernate.UserServiceImpl.getAllUsers(UserServiceImpl.java:33)
    at com.mk.hibernate.UserControllerImpl.getAllUsers(UserControllerImpl.java:27)
    at Main.main(Main.java:35)
Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: User is not mapped
    at org.hibernate.hql.internal.ast.util.SessionFactoryHelper.requireClassPersister(SessionFactoryHelper.java:189)
    at org.hibernate.hql.internal.ast.tree.FromElementFactory.addFromElement(FromElementFactory.java:109)
    at org.hibernate.hql.internal.ast.tree.FromClause.addFromElement(FromClause.java:95)
    at org.hibernate.hql.internal.ast.HqlSqlWalker.createFromElement(HqlSqlWalker.java:338)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElement(HqlSqlBaseWalker.java:3678)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElementList(HqlSqlBaseWalker.java:3567)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromClause(HqlSqlBaseWalker.java:708)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:564)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:301)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:249)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:278)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:206)
    ... 10 more
Feb 26, 2018 5:32:27 PM org.hibernate.engine.jdbc.connections.internal.DriverManagerConnectionProviderImpl stop
INFO: HHH000030: Cleaning up connection pool [jdbc:mysql://localhost:3306/demo]

我认为您误会了@Entity批注的name属性应该如何工作。

@Entity(name = "USERS")将User实体的hql名称映射到USERS,如果不存在任何值,则默认为classname。 因此,有效的hql查询将是"from USERS"而不是"from User"

您希望注释@Table(name = "USERS")定义实体的表名称。

尝试将映射注释更改为以下内容:

@Entity
@Table(name = "USERS")
public class User implements java.io.Serializable {
....
}

试试这个只是改变类名,它对我有用:

List<University> users = entityManager.createQuery(" from University").getResultList();
 System.out.println("University size = " + users.size());
 for (University university : users) {
    System.out.println("University Name = " + university.getName());
}

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM