Question

我一直试图让我的JPA（openJPA）应用程序使用非jta数据源与Tomcat一起工作。

我正在使用maven而没有IDE。只是vi在windows上，主要是为了了解事情是如何工作的并且聚集在一起

这是我得到的错误（我已经在SO上研究了类似的问题但无法找到修复程序）

Caused by: java.lang.RuntimeException: <openjpa-2.3.0-r422266:1540826 fatal user error> org.apache.openjpa.persistence.ArgumentException: The persistence provider is attempting to use properties in the persistence.xml file to resolve the data source. A Java Database Connectivity (JDBC) driver or data source class name must be specified in the openjpa.ConnectionDriverName or javax.persistence.jdbc.driver property. The following properties are available in the configuration: "org.apache.openjpa.jdbc.conf.JDBCConfigurationImpl@442ce698".

这是我的persistence.xml文件

    <?xml version="1.0" encoding="UTF-8"?>
<persistence 
    version="2.1" 
    xmlns="http://xmlns.jcp.org/xml/ns/persistence" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">

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

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

        <non-jta-data-source>java:/comp/env/jdbc/scoringdb</non-jta-data-source>

        <class>com.mantr.cricket.scorer.server.persistence.entities.Account</class>
        <class>com.mantr.cricket.scorer.server.persistence.entities.Delivery</class>
        <class>com.mantr.cricket.scorer.server.persistence.entities.DeliveryPK</class>
        <class>com.mantr.cricket.scorer.server.persistence.entities.Match</class>
        <class>com.mantr.cricket.scorer.server.persistence.entities.Player</class>
        <class>com.mantr.cricket.scorer.server.persistence.entities.Team</class>
    </persistence-unit>
</persistence>

这是Tomcat的context.xml文件

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

<Context>
<!--  JDBC Data source for the scoring database -->
    <Resource name="jdbc/scoringdb" 
              auth="Container"
              type="javax.sql.DataSource" 
              username="b888446f2789ce" 
              password="28fa221e"
              driverClassName="com.mysql.jdbc.Driver"
              url="jdbc:mysql://us-cdbr-east-06.cleardb.net/heroku_17744ef452a6a8d"
              maxActive="10" 
              maxIdle="4"/>
</Context>

Web.xml文件

    <?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
     xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
     version="2.5">



    <servlet>
        <servlet-name>cxf</servlet-name>
        <display-name>cxf</display-name>
        <description>Apache CXF Endpoint</description>
        <servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>cxf</servlet-name>
        <url-pattern>/services/*</url-pattern>
    </servlet-mapping>

    <resource-ref>
            <description>Scoring Database</description>
            <res-ref-name>jdbc/scoringdb</res-ref-name>
            <res-type>javax.sql.DataSource</res-type>
            <res-auth>Container</res-auth>
  </resource-ref>


</web-app>

当我尝试使用JNDI初始上下文打开连接时，这是有效的。所以JNDI数据源配置正确，我认为

Answer 1

请参阅我的Tomcat webapp的个人自动管理OpenJPA2.x包装器。查看jdbc名称，你可能会有一个额外的“/”字符？ https://github.com/Murmur/ScopedEntityManager/blob/master/example_webapp/webapp/WEB-INF/classes/META-INF/persistence.xml

<non-jta-data-source>java:/comp/env/jdbc/scoringdb</non-jta-data-source>
<non-jta-data-source>java:comp/env/jdbc/test</non-jta-data-source>

OpenJPA与Tomcat 8.无法创建EntityManager

1 个答案: