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java - 使用 hibernate 在 JPA 中配置 EntityManager

转载 作者:行者123 更新时间:2023-12-01 14:15:03 24 4
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我的应用程序的 META-INF 文件夹中有以下 persistence.xml:

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

<persistence version="2.0"
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">

<persistence-unit name="Hello" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>

<properties>
<property name="hibernate.show_sql" value="true" />
<property name="javax.persistence.jdbc.driver" value="oracle.jdbc.driver.OracleDriver" />
<property name="javax.persistence.jdbc.url" value="jdbc:oracle:thin:@my.db.com:1521:TEST"/>
<property name="javax.persistence.jdbc.user" value="newuser" />
<property name="javax.persistence.jdbc.password" value="password@123" />
<property name="hibernate.dialect" value="org.hibernate.dialect.Oracle10gDialect"/>
<property name="hibernate.hbm2ddl.auto" value="update"/>
</properties>
</persistence-unit>
</persistence>

web.xml 的上下文参数是:

<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/applicationContext.xml
/WEB-INF/rules-*.xml
</param-value>
</context-param>

在代码中,我正在执行以下操作:

EntityManagerFactory emf = Persistence
.createEntityManagerFactory("Hello");
EntityManager em = emf.createEntityManager();
RulesMasterTable rule = em.find(RulesMasterTable.class,
ruleName);

但异常(exception)的是:

javax.persistence.PersistenceException: No Persistence provider for EntityManager named Hello javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:69)javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:47)

为什么我的持久性单元没有注册?

最佳答案

我认为您不应该使用RESOURCE_LOCAL事务类型,请使用JTA。尝试在这里查看:Difference between a "jta-datasource" and a " resource-local " datasource?

关于java - 使用 hibernate 在 JPA 中配置 EntityManager,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18181262/

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