persistence.xml transaction-type =“ JTA”

Question

这个perisistence.xml文件正确吗，尤其是当我transaction-type="JTA" ？

<?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="CollDocPU" transaction-type="JTA">
<provider>org.apache.openjpa.persistence.PersistenceProviderImpl</provider>
<class>model.entity.StudentAddSolution</class>
<class>model.entity.Lecturer</class>
<class>model.entity.Solution</class>
<class>model.entity.Student</class>
<class>model.entity.Course</class>
<class>model.entity.File</class>
<class>model.entity.CourseHasLecturer</class>
<class>model.entity.Mail</class>
<class>model.entity.StudentAtCourse</class>
<class>model.entity.Roles</class>
<class>model.entity.Task</class>
<class>model.entity.User</class>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
<properties>
  <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:11080/myBase?zeroDateTimeBehavior=convertToNull"/>
  <property name="javax.persistence.jdbc.password" value="pass,"/>
  <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
  <property name="javax.persistence.jdbc.user" value="myBase"/>
  <property name="openjpa.jdbc.SynchronizeMappings" value="buildSchema(ForeignKeys=true)"/>
</properties>

Answer 1

您遇到错误了吗？

如果您使用的是JTA，建议您使用DataSource作为连接的提供程序。

如：

<?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="CollDocPU" transaction-type="JTA">
<provider>org.apache.openjpa.persistence.PersistenceProviderImpl</provider>
<jta-data-source>datasource</jta-data-source> 
<class>model.entity.StudentAddSolution</class>
<class>model.entity.Lecturer</class>
<class>model.entity.Solution</class>
<class>model.entity.Student</class>
<class>model.entity.Course</class>
<class>model.entity.File</class>
<class>model.entity.CourseHasLecturer</class>
<class>model.entity.Mail</class>
<class>model.entity.StudentAtCourse</class>
<class>model.entity.Roles</class>
<class>model.entity.Task</class>
<class>model.entity.User</class>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
</persistence-unit>

</persistence>

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问候

Answer 2

我的朋友，也许是另一个错误，请按照我告诉您的方法尝试此操作，只需通过System.out.println（em.toString（））打印EntityManager即可； 如果您正在获取某些东西，这是因为可以，您通过另一种方法获取的空指针可能是由程序逻辑引起的