openjpa

Using TomEE and open JPA, i get the following error: SEVERE: JAVA AGENT NOT INSTALLED

末鹿安然 提交于 2021-02-19 07:45:08
问题 I'm getting the following error using TomEE and JPA: SEVERE: JAVA AGENT NOT INSTALLED. The JPA Persistence Provider requested installation of a ClassFileTransformer which requires a JavaAgent. See http://openejb.apache.org/3.0/javaagent.html Is having a java agent required for openJPA to function properly? (Specifically is it needed for "openjpa.jdbc.SynchronizeMappings") If so, how do I install it properly in eclipse? 回答1: -javaagent:openejb-javaagent-3.0-beta-2.jar Add that to your VM

Hibernate结合JPA05

早过忘川 提交于 2020-12-05 07:00:59
一. JPA简介 JPA是Java Persistence API的简称,中文名Java持久层Api,是JDK1.5注解或者Xml描述对象-关系表的映射关系,并将运行期的实体类对象持久化Dao数据库中!注意的是,如果两种映射发生冲突的时候XML优先于注解的方式! 1. 什么是JPA? JPA由EJB 3.0软件专家组开发,作为JSR-220实现的一部分。但它又不限于EJB 3.0,你可以在Web应用、甚至桌面应用中使用。JPA的宗旨是为POJO提供持久化标准规范,由此可见,经过这几年的实践探索,能够脱离容器独立运行,方便开发和测试的理念已经深入人心了。Hibernate3.2+、TopLink 10.1.3以及 OpenJPA 都提供了JPA的实现。 JPA的总体思想和现有Hibernate、TopLink、JDO等ORM框架大体一致。总的来说,JPA包括以下3方面的技术: ORM映射元数据 JPA支持XML和 JDK 5.0注解两种元数据的形式,元数据描述对象和表之间的映射关系,框架据此将实体对象持久化到数据库表中; API 用来操作实体对象,执行CRUD操作,框架在后台替代我们完成所有的事情,开发者从繁琐的JDBC和SQL代码中解脱出来。 查询语言 这是持久化操作中很重要的一个方面,通过面向对象而非面向数据库的查询语言查询数据,避免程序的SQL语句紧密耦合。 2.

Why do backslashes need to be escaped in JPA native queries?

百般思念 提交于 2020-06-15 18:53:27
问题 In JPA, when executing a native query, I have to escape backslash characters to double backslash. E.g. when having a where clause like UPPER(app.NAME) like UPPER(?) escape '\' then I need to create a string which contains: UPPER(app.NAME) like UPPER(?) escape '\\' Then I run a native query using openJPA 2.2.2: final EntityManager entityManager = EntityManagerFinder.getEntityManager(); final Query query = entityManager.createNativeQuery(sql, Response.class); The SQL statement is read from XML

关于:Table '项目名称..hibernate_sequence' doesn't exist的解决方法

只愿长相守 提交于 2020-03-25 15:02:38
3 月,跳不动了?>>> 解决方案: 这个是主键自增长策略问题。 将ID生成略组改成@GeneratedValue(strategy = GenerationType.IDENTITY). 关于主键策略 @GeneratedValue:主键的产生策略,通过strategy属性指定。 主键产生策略通过GenerationType来指定。GenerationType是一个枚举,它定义了主键产生策略的类型。 1、AUTO自动选择一个最适合底层数据库的主键生成策略。如MySQL会自动对应auto increment。这个是默认选项,即如果只写@GeneratedValue,等价于@GeneratedValue (strategy=GenerationType.AUTO)。 2、IDENTITY 表自增长字段,Oracle不支持这种方式。 3、SEQUENCE 通过序列产生主键,MySQL不支持这种方式。 4、TABLE 通过表产生主键,框架借由表模拟序列产生主键,使用该策略可以使应用更易于数据库移植。不同的JPA实现商生成的表名是不同的,如 OpenJPA生成openjpa_sequence_table表,Hibernate生成一个hibernate_sequences表,而TopLink则生成sequence表。这些表都具有一个序列名和对应值两个字段,如SEQ_NAME和SEQ

ZhaoWei-2020-02-12

五迷三道 提交于 2020-02-28 04:23:53
起源 JPA由EJB 3.0软件专家组开发,作为JSR-220实现的一部分。但它不囿于EJB 3.0,你可以在Web应用、甚至桌面应用中使用。JPA的宗旨是为POJO提供持久化标准规范,由此可见,经过这几年的实践探索,能够脱离容器独立运行,方便开发和测试的理念已经深入人心了。目前Hibernate 3.2、TopLink 10.1.3以及OpenJPA都提供了JPA的实现 内容 JPA(Java Persistence API)是SUN官方推出的Java持久化规范,它为Java开发人员提供了一种对象/关联映射工具来管理Java应用中的关系数据。它的出现主要是为了简化现有的持久化开发工作和整合ORM技术,结束现在Hibernate,TopLink,JDO等ORM框架各自为营的局面。值得注意的是,JPA是在充分吸收了现有Hibernate,TopLink,JDO 等ORM框架的基础上发展而来的,具有易于使用,伸缩性强等优点。从开发社区的反映上看,JPA手动极大的支持和赞扬,其中就包括了Spring与EJB3.0的开发团队,着眼未来几年的技术走向,JPA作为ORM领域的标准化者的目标应该不难实现。 JPA包括一下三方面的技术 *ORM映射元数据 JPA支持XML和JDK5.0注释两种元数据形式,元数据描述对象和表之间的映射关系,框架据此将实体对象持久化到数据库表中。

java.lang.ClassCastException: [B cannot be cast to java.lang.String

混江龙づ霸主 提交于 2020-02-27 22:40:03
问题 I have written an entitity class with Field LoginId and Password. Iam encrypting the passwrd and stoiring it in the db using the AES_ENCRYPT. I want to retrive only the password which is decrypted. so, im using AES_DECRYPT using NAtiveQueryis in OPen JPA 2.0. Query i have written is : Query q = em.createNativeQuery("select AES_DECRYPT(l.password,?2) from loginDetails l where l.loginID = ?1"); q.setParameter(1, loginId); q.setParameter(2, getKey()); String s = q.getSingleResult(); But im

configure openjpa on to spring boot

帅比萌擦擦* 提交于 2020-01-23 13:28:25
问题 Im trying to change the default JPA implementation of Hibernate to OpenJPA on Spring boot. Ive searched on google but there is not much on how to configure openJPA to Spring boot. Any advice would be helpful. Thanks 回答1: @Configuration public class OpenJPAConfig extends JpaBaseConfiguration { protected OpenJPAConfig(DataSource dataSource, JpaProperties properties, ObjectProvider<JtaTransactionManager> jtaTransactionManager, ObjectProvider<TransactionManagerCustomizers>

configure openjpa on to spring boot

喜欢而已 提交于 2020-01-23 13:28:10
问题 Im trying to change the default JPA implementation of Hibernate to OpenJPA on Spring boot. Ive searched on google but there is not much on how to configure openJPA to Spring boot. Any advice would be helpful. Thanks 回答1: @Configuration public class OpenJPAConfig extends JpaBaseConfiguration { protected OpenJPAConfig(DataSource dataSource, JpaProperties properties, ObjectProvider<JtaTransactionManager> jtaTransactionManager, ObjectProvider<TransactionManagerCustomizers>

JPA Entity mapping for join columns

瘦欲@ 提交于 2020-01-16 08:59:13
问题 I have three tables EmployeeDepartment, EmployeeGroup and EmpplyeeDetails Table. EmployeeDepartment table has primary key departmentId and a column groupId, EmployeeGroupTable has primary key groupid which should be generated from databse sequence GroupIdGenerator. EmployeeDetails have two primary keys as groupid and employeeid. Groupid should be same as of the previous table These values in all table should insert in one transaction. Can you help me with correct JAP Entity mapping? I already

OpenJPA 2.1.1 - Cannot find the declaration of element 'persistence'

若如初见. 提交于 2020-01-14 10:32:39
问题 I just downloaded http://www.apache.org/dyn/closer.cgi/openejb/4.0.0-beta-1/apache-tomee-1.0.0-beta-1-webprofile.zip to use OpenEJB with OpenJPA2.1.1. I can't get my persistence.xml working. The top of the stack trace: org.xml.sax.SAXException: file:/D:/Workspaces/sandbox/tomcat_ejb_jpa2_tomEE/.metadata/.plugins/org.eclipse.wst.server.core/tmp0/wtpwebapps/testEE/WEB-INF/classes/META-INF/persistence.xml [Location: Line: 2, C: 248]: org.xml.sax.SAXParseException: cvc-elt.1: Cannot find the