阵列，在Hibernate和PostgreSQL用户类型 - ＆GT; MappingException [英] Array with UserType in Hibernate and PostgreSQL --> MappingException
问题描述
我想写实现用户类型来应对休眠/ JPA数组类。
我用下面的帖子
<一href=\"http://stackoverflow.com/questions/1647583/mapping-a-postgres-array-with-hibernate/1647660#1647660\">mapping一个Postgres阵列休眠和<一个href=\"http://stackoverflow.com/questions/5393334/hibernate-jpa-hsql-how-to-create-a-dialect-mapping-for-user-type-array/5412235#5412235\">Hibernate/JPA/HSQL ：如何创建一个方言映射用户类型ARRAY 建立一个解决方案，但我不能让它工作。
我创建了一个新的Spring Roo项目只是为了测试它。下面是不同的文件（所有Java类都位于包检测）：
I am trying to write a class implementing UserType to deal with arrays in Hibernate/JPA. I used the following posts mapping a postgres array with hibernate and Hibernate/JPA/HSQL : How to create a Dialect mapping for User Type ARRAY to build a solution but I cannot get it to work. I created a new Spring Roo project just to test it. Here are the different files (all java classes are located in the package test):
-
的persistence.xml
persistence.xml
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.0" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
<persistence-unit name="persistenceUnit" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<properties>
<!-- <property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQLDialect"/> -->
<property name="hibernate.dialect" value="test.PostgreSQLDialectArray"/>
<!-- value="create" to build a new database on each run; value="update" to modify an existing database; value="create-drop" means the same as "create" but also drops tables when Hibernate closes; value="validate" makes no changes to the database -->
<property name="hibernate.hbm2ddl.auto" value="create"/>
<property name="hibernate.ejb.naming_strategy" value="org.hibernate.cfg.ImprovedNamingStrategy"/>
<property name="hibernate.connection.charSet" value="UTF-8"/>
<!-- Uncomment the following two properties for JBoss only -->
<!-- property name="hibernate.validator.apply_to_ddl" value="false" /-->
<!-- property name="hibernate.validator.autoregister_listeners" value="false" /-->
</properties>
</persistence-unit>
</persistence>
TestArray.java
TestArray.java
package test;
import java.math.BigInteger;
import java.security.SecureRandom;
import org.springframework.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;
import org.springframework.roo.addon.javabean.RooJavaBean;
import org.springframework.roo.addon.serializable.RooSerializable;
import org.springframework.roo.addon.tostring.RooToString;
@RooJavaBean
@RooToString
@RooSerializable
public class TestArray {
private static final long serialVersionUID = 1L;
private SecureRandom random = new SecureRandom();
public String nextSessionId()
{
return new BigInteger(130, random).toString(32);
}
public static void main(String[] args) {
ApplicationContext context;
context = new ClassPathXmlApplicationContext("classpath:META-INF/spring/applicationContext.xml");
int[] array = new int[1428];
TestArray test = new TestArray();
Blabla blabla = new Blabla();
int nb = 1428;
for(int i = 0 ; i < nb ; i++)
array[i] = test.random.nextInt();
// blabla.setTest(array);
// blabla.persist();
// System.out.println(Arrays.toString(blabla.getTest()));
System.out.println(java.sql.Types.ARRAY);
System.out.println("Done");
}
}
Blabla.java
Blabla.java
package test;
import org.hibernate.annotations.Type;
import org.springframework.roo.addon.entity.RooEntity;
import org.springframework.roo.addon.javabean.RooJavaBean;
import org.springframework.roo.addon.tostring.RooToString;
@RooJavaBean
@RooToString
@RooEntity
public class Blabla {
@Type(type = "test.IntArrayUserType")
private int[] array;
}
PostgreSQLDialectArray
PostgreSQLDialectArray
package test;
import java.sql.Types;
public class PostgreSQLDialectArray extends org.hibernate.dialect.PostgreSQLDialect{
public PostgreSQLDialectArray() {
super();
registerHibernateType(Types.ARRAY, "array");
}
}
IntArrayUserType.java（基本上比<一个相同href=\"http://stackoverflow.com/questions/1647583/mapping-a-postgres-array-with-hibernate/1647660#1647660\">mapping一个Postgres阵列休眠）
package test;
import java.io.Serializable;
import java.sql.Array;
import java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import org.hibernate.HibernateException;
import org.hibernate.usertype.UserType;
public class IntArrayUserType implements UserType {
protected static final int SQLTYPE = java.sql.Types.ARRAY;
private int[] toPrimitive(Integer[] array){
int[] a = new int[array.length];
for(int i = 0 ; i < array.length ; i++)
a[i] = array[i];
return a;
}
private Integer[] toObject(int[] array){
Integer[] a = new Integer[array.length];
for(int i = 0 ; i < array.length ; i++)
a[i] = array[i];
return a;
}
@Override
public Object nullSafeGet(final ResultSet rs, final String[] names, final Object owner) throws HibernateException, SQLException {
Array array = rs.getArray(names[0]);
Integer[] javaArray = (Integer[]) array.getArray();
return toPrimitive(javaArray);
}
@Override
public void nullSafeSet(final PreparedStatement statement, final Object object, final int i) throws HibernateException, SQLException {
System.out.println("test null safe set...");
Connection connection = statement.getConnection();
int[] castObject = (int[]) object;
Integer[] integers = toObject(castObject);
Array array = connection.createArrayOf("integer", integers);
statement.setArray(i, array);
System.out.println("test null safe set...");
}
@Override
public Object assemble(final Serializable cached, final Object owner) throws HibernateException {
return cached;
}
@Override
public Object deepCopy(final Object o) throws HibernateException {
return o == null ? null : ((int[]) o).clone();
}
@Override
public Serializable disassemble(final Object o) throws HibernateException {
return (Serializable) o;
}
@Override
public boolean equals(final Object x, final Object y) throws HibernateException {
return x == null ? y == null : x.equals(y);
}
@Override
public int hashCode(final Object o) throws HibernateException {
return o == null ? 0 : o.hashCode();
}
@Override
public boolean isMutable() {
return false;
}
@Override
public Object replace(final Object original, final Object target, final Object owner) throws HibernateException {
return original;
}
@Override
public Class<int[]> returnedClass() {
return int[].class;
}
@Override
public int[] sqlTypes() {
return new int[] { SQLTYPE };
}
}
而现在的堆栈跟踪：
Exception in thread "main" org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'transactionManager' defined in class path resource [META-INF/spring/applicationContext.xml]: Cannot resolve reference to bean 'entityManagerFactory' while setting bean property 'entityManagerFactory'; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [META-INF/spring/applicationContext.xml]: Invocation of init method failed; nested exception is javax.persistence.PersistenceException: [PersistenceUnit: persistenceUnit] Unable to build EntityManagerFactory
at org.springframework.beans.factory.support.BeanDefinitionValueResolver.resolveReference(BeanDefinitionValueResolver.java:328)
at org.springframework.beans.factory.support.BeanDefinitionValueResolver.resolveValueIfNecessary(BeanDefinitionValueResolver.java:106)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.applyPropertyValues(AbstractAutowireCapableBeanFactory.java:1325)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.populateBean(AbstractAutowireCapableBeanFactory.java:1086)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:517)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:456)
at org.springframework.beans.factory.support.AbstractBeanFactory$1.getObject(AbstractBeanFactory.java:291)
at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:222)
at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:288)
at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:190)
at org.springframework.beans.factory.support.DefaultListableBeanFactory.preInstantiateSingletons(DefaultListableBeanFactory.java:580)
at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:895)
at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:425)
at org.springframework.context.support.ClassPathXmlApplicationContext.<init>(ClassPathXmlApplicationContext.java:139)
at org.springframework.context.support.ClassPathXmlApplicationContext.<init>(ClassPathXmlApplicationContext.java:83)
at test.TestArray.main(TestArray.java:29)
Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [META-INF/spring/applicationContext.xml]: Invocation of init method failed; nested exception is javax.persistence.PersistenceException: [PersistenceUnit: persistenceUnit] Unable to build EntityManagerFactory
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1420)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:519)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:456)
at org.springframework.beans.factory.support.AbstractBeanFactory$1.getObject(AbstractBeanFactory.java:291)
at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:222)
at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:288)
at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:190)
at org.springframework.beans.factory.support.BeanDefinitionValueResolver.resolveReference(BeanDefinitionValueResolver.java:322)
... 15 more
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: persistenceUnit] Unable to build EntityManagerFactory
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:915)
at org.hibernate.ejb.HibernatePersistence.createContainerEntityManagerFactory(HibernatePersistence.java:74)
at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:225)
at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:308)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1477)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1417)
... 22 more
Caused by: org.hibernate.MappingException: No Dialect mapping for JDBC type: 2003
at org.hibernate.dialect.TypeNames.get(TypeNames.java:77)
at org.hibernate.dialect.TypeNames.get(TypeNames.java:100)
at org.hibernate.dialect.Dialect.getTypeName(Dialect.java:296)
at org.hibernate.mapping.Column.getSqlType(Column.java:208)
at org.hibernate.mapping.Table.sqlCreateString(Table.java:418)
at org.hibernate.cfg.Configuration.generateSchemaCreationScript(Configuration.java:1099)
at org.hibernate.tool.hbm2ddl.SchemaExport.<init>(SchemaExport.java:106)
at org.hibernate.impl.SessionFactoryImpl.<init>(SessionFactoryImpl.java:372)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1872)
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:906)
... 27 more
所以我想这是我创建的话完全不使用或者是错误的，但我不知道为什么。我认为@Type注释应该做的映射，但我看到了一些标签，但如果需要他们在这个例子中，我不知道，如果是这样，我在哪里了呢？它已经两天我坚持这个问题，我越来越绝望。你能不能帮我调试这个计划？请。
非常感谢你提前。
So I guess that the dialect that I created is not used at all or is wrong but I don't know why. I think that the @Type annotation should do the mapping but I have seen some tags but I do not know if they are needed in this example and if they are, where do I had them? It's been two days that I am stuck with this problem and I am getting desperate. Could you help me debug this program? Please. Thank you very much in advance.
编辑0：
persistence.xml文件似乎找到正确的话，但在功能getTypeName（2003）抛出上述操作的registerHibernateType（Types.ARRAY，阵列）之后的错误。顺便说一句，我使用Hibernate的3.6.4.Final和PostgreSQL 8.4-702.jdbc3。
The persistence.xml file seem to find the correct dialect but the function getTypeName(2003) throws the errors above after doing registerHibernateType(Types.ARRAY, "array"). BTW, I am using Hibernate 3.6.4.Final and postgresql 8.4-702.jdbc3.
编辑1：
我添加了以下行PostgreSQLDialectArray构造：
I added the following line to PostgreSQLDialectArray constructor:
registerColumnType(Types.ARRAY, "integer[$l]" );
这似乎部分地解决这个问题。不过，现在我得到另一个错误：
which seems to partially solve the problem. However, now I get another error:
2013-01-09 11:14:30,281 [main] ERROR org.hibernate.tool.hbm2ddl.SchemaExport - Unsuccessful: create table blabla (id int8 not null, array int[255], name varchar(255), test int4 not null, version int4, primary key (id))
2013-01-09 11:14:30,282 [main] ERROR org.hibernate.tool.hbm2ddl.SchemaExport - ERREUR: erreur de syntaxe sur ou près de « array »
Position: 40
显然，休眠还是不知道如何创建一个表，在一个数组...
Apparently, hibernate still does not know how to create a table with an array in it...
编辑2：
看来PostgreSQL的不喜欢的事实，我的专栏被称为阵。我改变，它的工作。该表是由Hibernate整数数组创建。
It seems that postgresql didn't like the fact that my column was called "array". I change that and it worked. The table is created by hibernate with an array of integer.
但我不能在它与Hibernate保存，因为在用户类型执行问题的数组。显然，阵列的创建失败createArrayOf。我读关于此事告诉访问底层连接，而不是包装一些线程。我想我会打开一个新的线程链接到这一个，因为这个问题是完全不同的。
BUT I cannot save any array in it with hibernate because of a problem in the UserType implementation. Apparently, the creation of the array fails with createArrayOf. I am reading some threads on this matter telling to access the underlying connection instead of the wrapper. I think I am going to open a new thread linking to this one because this problem is quite different.
该堆栈跟踪：
Exception in thread "main" java.lang.AbstractMethodError: org.apache.commons.dbcp.PoolingDataSource$PoolGuardConnectionWrapper.createArrayOf(Ljava/lang/String;[Ljava/lang/Object;)Ljava/sql/Array;
at test.IntArrayUserType.nullSafeSet(IntArrayUserType.java:59)
at org.hibernate.type.CustomType.nullSafeSet(CustomType.java:140)
at org.hibernate.persister.entity.AbstractEntityPersister.dehydrate(AbstractEntityPersister.java:2184)
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:2430)
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:2874)
at org.hibernate.action.EntityInsertAction.execute(EntityInsertAction.java:79)
at org.hibernate.engine.ActionQueue.execute(ActionQueue.java:273)
at org.hibernate.engine.ActionQueue.executeActions(ActionQueue.java:265)
at org.hibernate.engine.ActionQueue.executeActions(ActionQueue.java:184)
at org.hibernate.event.def.AbstractFlushingEventListener.performExecutions(AbstractFlushingEventListener.java:321)
at org.hibernate.event.def.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:51)
at org.hibernate.impl.SessionImpl.flush(SessionImpl.java:1216)
at org.hibernate.impl.SessionImpl.managedFlush(SessionImpl.java:383)
at org.hibernate.transaction.JDBCTransaction.commit(JDBCTransaction.java:133)
at org.hibernate.ejb.TransactionImpl.commit(TransactionImpl.java:76)
at org.springframework.orm.jpa.JpaTransactionManager.doCommit(JpaTransactionManager.java:467)
at org.springframework.transaction.support.AbstractPlatformTransactionManager.processCommit(AbstractPlatformTransactionManager.java:754)
at org.springframework.transaction.support.AbstractPlatformTransactionManager.commit(AbstractPlatformTransactionManager.java:723)
at org.springframework.transaction.interceptor.TransactionAspectSupport.commitTransactionAfterReturning(TransactionAspectSupport.java:393)
at org.springframework.transaction.aspectj.AbstractTransactionAspect.ajc$afterReturning$org_springframework_transaction_aspectj_AbstractTransactionAspect$3$2a73e96c(AbstractTransactionAspect.aj:78)
at test.Blabla_Roo_Entity.ajc$interMethod$test_Blabla_Roo_Entity$test_Blabla$persist(Blabla_Roo_Entity.aj:56)
at test.Blabla.persist(Blabla.java:1)
at test.Blabla_Roo_Entity.ajc$interMethodDispatch1$test_Blabla_Roo_Entity$test_Blabla$persist(Blabla_Roo_Entity.aj)
at test.TestArray.main(TestArray.java:39)
修改3：
最后，下面的修改之后，对于整型数组的用户类型如下：
Finally, after the following modifications, the UserType for integer arrays works:
-
在applicationContext.xml中加入这一行：
Add this line in applicationContext.xml in:
＆LT; bean类=org.apache.commons.dbcp.BasicDataSource消灭法=关闭ID =数据源＆GT; 结果
... 结果
＆LT;属性名=accessToUnderlyingConnectionAllowedVALUE =真/＆GT; 结果
＆LT; /豆＆GT; 搜索
从IntArrayUserType修改nullSafeSet
Modify the nullSafeSet from IntArrayUserType
@Override
public void nullSafeSet(final PreparedStatement statement, final Object object, final int i) throws HibernateException, SQLException {
Connection connection = statement.getConnection();
int[] castObject = (int[]) object;
Integer[] integers = toObject(castObject);
Connection conn = ((DelegatingConnection) connection).getInnermostDelegate();
Array array = conn.createArrayOf("integer", integers);
statement.setArray(i, array);
}
但仍然得到从表布拉布拉所有条目时出现问题：
该功能findAllBlablas不能正常工作，只返回第一个条目...
BUT there is still a problem when getting all the entries from the table blabla: The function findAllBlablas does not work properly and only returns the first entry...
修改4：
在事实上，它的工作很好，但Eclipse控制台无法打印所有数据。这就是全部！
In fact, it worked great but the eclipse console was not able to print all data. That's all!
推荐答案
我回答我所有的编辑我自己的问题。我还张贴此评论的情况下，其他人被困像我这样他们就可以看到这个问题的回答。然而，随着findAllBlablas最后问题仍然存在。如果我发现一个解决方法，我会重新编辑我的问题。
I answered my own question in all my edits. I still post this comment in case other people get stuck like me so they can see that this problem was answered. However, the last problem with findAllBlablas remains. If I find a fix, I'll edit my question again.
这篇关于阵列，在Hibernate和PostgreSQL用户类型 - ＆GT; MappingException的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
