关于这个问题肯定有很多问题,我已经读了一些,但答案仍然没有找到答案.我是JPA的新手,我只是想测试一个简单的应用程序,看看我是否可以正确配置它.它是一个独立的应用程序,意味着它不会与Web服务器或任何东西一起运行.实体类看起来像:
@Entity
public class Person{
@Id
private String userID = null;
@Transient
private UserState userState = null;
private String email = null;
private String name = null;
public Person(){
userID = null;
email = null;
name = null;
userState = null;
}
public String getUserID() {
return userID;
}
public void setUserID(String userID) {
this.userID = userID;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public UserState getUserState() {
return userState;
}
public void setUserState(UserState userState) {
this.userState = userState;
}
}
主要的:
public class PersistenceTest {
public static void main(String[] args) {
System.out.println("creating person");
Person p = new Person();
p.setUserID("GregR");
p.setEmail("Gregory@company.de");
p.setUserState(UserState.ACTIVE);
System.out.println("done creating person GregR");
EntityManagerFactory factory = Persistence.createEntityManagerFactory(PersonService.PERSISTENCE_UNIT_NAME);
System.out.println("factory initialized");
EntityManager manager = factory.createEntityManager();
System.out.println("EntityManager initialized");
PersonService service = new PersonService(manager);
System.out.println("service initialized");
System.out.println("Beginning transaction");
manager.getTransaction().begin();
System.out.println("Transaction begun");
System.out.println("attempting to persist person");
service.persistEntity(p);
manager.getTransaction().commit();
System.out.println("person persisted");
System.out.println("beginning cleanup");
manager.close();
factory.close();
System.out.println("Cleanup has completed");
}
}
配置:
org.apache.openjpa.persistence.PersistenceProviderImpl
de.hol.persistable.entities.Person
控制台打印输出:
creating person
done creating person GReeder
factory initialized
47 PersonService INFO [main] openjpa.Runtime - Starting OpenJPA 2.3.0
110 PersonService INFO [main] openjpa.jdbc.JDBC - Using dictionary class "org.apache.openjpa.jdbc.sql.MySQLDictionary".
306 PersonService INFO [main] openjpa.jdbc.JDBC - Connected to MySQL version 5.5 using JDBC driver MySQL-AB JDBC Driver version mysql-connector-java-5.0.8 ( Revision: ${svn.Revision} ).
Exception in thread "main" <1540826 nonfatal user error> org.apache.openjpa.persistence.ArgumentException: This configuration disallows runtime optimization, but the following listed types were not enhanced at build time or at class load time with a javaagent: "1540826>
de.hol.persistable.entities.Person".
at org.apache.openjpa.enhance.ManagedClassSubclasser.prepareUnenhancedClasses(ManagedClassSubclasser.java:115)
at org.apache.openjpa.kernel.AbstractBrokerFactory.loadPersistentTypes(AbstractBrokerFactory.java:312)
at org.apache.openjpa.kernel.AbstractBrokerFactory.initializeBroker(AbstractBrokerFactory.java:236)
at org.apache.openjpa.kernel.AbstractBrokerFactory.newBroker(AbstractBrokerFactory.java:212)
at org.apache.openjpa.kernel.DelegatingBrokerFactory.newBroker(DelegatingBrokerFactory.java:155)
at org.apache.openjpa.persistence.EntityManagerFactoryImpl.createEntityManager(EntityManagerFactoryImpl.java:226)
at org.apache.openjpa.persistence.EntityManagerFactoryImpl.createEntityManager(EntityManagerFactoryImpl.java:153)
at org.apache.openjpa.persistence.EntityManagerFactoryImpl.createEntityManager(EntityManagerFactoryImpl.java:59)
at de.hol.persistable.PersistenceTest.main(PersistenceTest.java:24)
我的问题
我猜主要的问题是,我做错了什么.我对此非常陌生,我正试图让这个独立的应用程序工作,以便我可以扩展它以供现实世界使用.
2.我是否缺少persistence.xml文件以外的其他一些配置?
3.为独立应用程序解决此错误的最简单方法是什么?
提前谢谢了!
解决方法:
我看到你有主类,因此我假设你在Java SE环境中使用它.使其工作的最简单方法是在命令行中定义-javaagent,如下所示:
java -jar myJAR.jar -javaagent:openjpa-all-2.3.0.jar
也可以从Eclipse:Run-> Run Configurations->在“Java Applications”中找到您的应用程序 – > Arguments-> VM arguments-> add
-javaagent:/full/path/to/openjpa-all-2.3.0.jar
标签:openjpa,java,mysql,jpa
来源: https://codeday.me/bug/20190728/1561773.html