我试图将我的PostgreSQL db与OpenJPA和Spring连接起来。如何连接OpenJPA，Spring和PostgreSQL？

这是我的Spring上下文文件。它包含MVC部分和TX。

<beans xmlns="http://www.springframework.org/schema/beans" 
     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
     xmlns:ctx="http://www.springframework.org/schema/context" 
     xmlns:tx="http://www.springframework.org/schema/tx" 
     xsi:schemaLocation="http://www.springframework.org/schema/beans 
     http://www.springframework.org/schema/beans/spring-beans.xsd 
     http://www.springframework.org/schema/context 
     http://www.springframework.org/schema/context/spring-context.xsd 
     http://www.springframework.org/schema/tx 
     http://www.springframework.org/schema/tx/spring-tx.xsd"> 

    <ctx:annotation-config/> 
    <ctx:component-scan base-package="de.alex.web.pag"/> 
    <tx:annotation-driven transaction-manager="transactionManager"/> 

    <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver"> 
     <property name="prefix" value="/WEB-INF/pages/"/> 
     <property name="suffix" value=".jsp"/> 
    </bean> 

    <bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager"> 
     <property name="entityManagerFactory" ref="entityManagerFactory"/> 
    </bean> 

    <bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean"> 
     <property name="persistenceUnitName" value="pagPU"/> 
     <property name="persistenceXmlLocation" value="/META-INF/persistence.xml"/> 
     <property name="jpaVendorAdapter" ref="openJpaVendorAdapter"/> 
     <property name="dataSource" ref="dataSource"/> 
    </bean> 

    <bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource"> 
     <property name="driverClassName" value="org.postgresql.Driver"/> 
     <property name="url" value="jdbc:postgresql://localhost:5432/box"/> 
     <property name="username" value="postgres"/> 
     <property name="password" value="bob"/> 
    </bean> 

    <bean id="openJpaVendorAdapter" class="org.springframework.orm.jpa.vendor.OpenJpaVendorAdapter"> 
     <property name="showSql" value="true"/> 
     <property name="generateDdl" value="true"/> 
     <property name="database" value="POSTGRESQL"/> 
     <property name="databasePlatform" value="org.apache.openjpa.jdbc.sql.PostgresDictionary"/> 
    </bean> 
</beans>

这是我在/properties/META-INF/文件夹中的persistence.xml：

<persistence xmlns="http://java.sun.com/xml/ns/persistence" 
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
      xsi:schemaLocation="http://java.sun.com/xml/ns/persistence 
      http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd" 
      version="2.0"> 

    <persistence-unit name="pagPU" transaction-type="RESOURCE_LOCAL"> 

     <provider>org.apache.openjpa.persistence.PersistenceProviderImpl</provider> 

     <class>de.alex.web.pag.entity.Category</class> 
     <class>de.alex.web.pag.entity.Order</class> 
     <class>de.alex.web.pag.entity.Product</class> 

     <properties> 
      <property name="openjpa.ConnectionURL" value="jdbc:postgresql://localhost:5432/box"/> 
      <property name="openjpa.jdbc.DBDictionary" value="postgres"/> 
      <property name="openjpa.ConnectionDriverName" value="org.postgresql.Driver"/> 
      <property name="openjpa.ConnectionUserName" value="postgres"/> 
      <property name="openjpa.ConnectionPassword" value="bob"/> 
      <property name="openjpa.Log" value="SQL=TRACE"/> 
     </properties> 
    </persistence-unit> 
</persistence>

当我尝试在我的IDE中运行这个简单的脚本，我收到此异常：

[2014-09-26 17:16:43] java.lang.RuntimeException: <openjpa-2.3.0-r422266:1540826 nonfatal user error> org.apache.openjpa.persistence.ArgumentException: An error occurred while parsing the query filter "select c from Category c where c.id = 1". Error message: The name "Category" is not a recognized entity or identifier. Known entity names: [] 
    at org.apache.openjpa.kernel.exps.AbstractExpressionBuilder.parseException(AbstractExpressionBuilder.java:119) 
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder.getClassMetaData(JPQLExpressionBuilder.java:197) 
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder.resolveClassMetaData(JPQLExpressionBuilder.java:167) 
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder.getCandidateMetaData(JPQLExpressionBuilder.java:242) 
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder.getCandidateMetaData(JPQLExpressionBuilder.java:212) 
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder.getCandidateType(JPQLExpressionBuilder.java:205) 
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder.access$200(JPQLExpressionBuilder.java:80) 
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder$ParsedJPQL.populate(JPQLExpressionBuilder.java:2428) 
    at org.apache.openjpa.kernel.jpql.JPQLParser.populate(JPQLParser.java:61) 
    at org.apache.openjpa.kernel.ExpressionStoreQuery.populateFromCompilation(ExpressionStoreQuery.java:162) 
    at org.apache.openjpa.kernel.QueryImpl.newCompilation(QueryImpl.java:673) 
    at org.apache.openjpa.kernel.QueryImpl.compilationFromCache(QueryImpl.java:654) 
    at org.apache.openjpa.kernel.QueryImpl.compileForCompilation(QueryImpl.java:620) 
    at org.apache.openjpa.kernel.QueryImpl.compileForExecutor(QueryImpl.java:682) 
    at org.apache.openjpa.kernel.QueryImpl.compile(QueryImpl.java:589) 
    at org.apache.openjpa.persistence.EntityManagerImpl.createQuery(EntityManagerImpl.java:997) 
    at org.apache.openjpa.persistence.EntityManagerImpl.createQuery(EntityManagerImpl.java:979) 
    at org.apache.openjpa.persistence.EntityManagerImpl.createQuery(EntityManagerImpl.java:102) 
    in RemoteEntityManagerImpl.createQuery(RemoteEntityManagerImpl.java:39) 
    in RemoteUtil.executeWithClassLoader(RemoteUtil.java:167) 
    in RemoteUtil$2$1.invoke(RemoteUtil.java:102) 
    at com.sun.proxy.$Proxy151.createQuery(Unknown Source) 
    in JpaEngine.createQuery(JpaEngine.java:104)

看来我当我配置Spring上下文或JPA上下文时错过了重要的东西。

UPDATE

我的实体类：

@Entity 
@Table(name = "category") 
public class Category { 
    @Id 
    @SequenceGenerator(name = "pk_seq", sequenceName = "category_table_id_seq", allocationSize = 1) 
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "pk_seq") 
    private Integer id; 

    @Column(name = "name") 
    private String name; 

    public Category() { 
    } 
    // getters and setters ommited 
}

UPD2

现在我得到这个异常：

org.springframework.transaction.CannotCreateTransactionException: Could not open JPA EntityManager for transaction; nested exception is <openjpa-2.3.0-r422266:1540826 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: " 
de.alex.web.pag.entity.Category 
de.alex.web.pag.entity.Product 
de.alex.web.pag.entity.Order".

来源

2014-09-26 Alex

你有@Entity标注为模型类？ – maszter 2014-09-26 15:46:06

是的。我更新了这个问题。 – Alex 2014-09-26 15:47:49

你说你persistence.xml文件在'/ properties/META-INF /'中，而在Spring配置文件中，你声明它在'/ META-INF/persistence.xml'。你确定它是否正确？ – 2014-09-26 17:03:38

你这里的问题是，默认情况Ope宇宙nJPA使用字节码（编译时间）来增强持久化实体，而不是说例如Hibernate的运行时代理方法。

你因此需要：

在编译时提高了持久化类。参见：http://openjpa.apache.org/entity-enhancement.html。如果您正在使用Maven，请在您的POM中配置插件。如果您还在使用Eclipse，则可以配置该插件以节省每次对持久性类进行更改时必须运行完整“mvn clean install”的情况。

http://openjpa.apache.org/openjpaeclipseenhancementbuilder.html

或

启用运行时增强：http://openjpa.apache.org/runtime-enhancement.html

来源

2014-09-26 22:11:05

如何连接OpenJPA，Spring和PostgreSQL？

UPDATE

UPD2

回答

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