问题
I have a fairly standard Java EE6 web application using JPA 2 with dependency injection connecting to a MySQL database and everything is working fine. What I would like to do now is have this application interact with the databases of other applications we have installed at a clients site - essentially acting as a single point of control for our other application installs.
What I'm struggling with is how best to perform the interaction with the other databases. Ideally I would like to create an EntityManager for each install and interact using JPA but I can't see any way to set this up. I may, for example, have 5 installs (and therefore databases) of one application type and the master control application won't know about the other installs until runtime. This seems to preclude using dependency injection of an EntityManager and all the automatic transaction demacation etc etc. The alternative option is to just create a DataSource and do the interactions manually. While flexible this clearly requires a lot more effort.
So, my question really is how do I best tackle this problem?
回答1:
I'm also looking into this, and so far I have found the following blog post that describes a way to do it http://ayushsuman.blogspot.com/2010/06/configure-jpa-during-run-time-dynamic.html :
Removed all your database properties from persistance.xml
<persistence>
<persistence-unit name="jpablogPUnit" transaction-type="RESOURCE_LOCAL">
<class>com.suman.Company</class>
</persistence-unit>
</persistence>
Changed your java file where you are configuring entityManager, in our case TestApplication.java
package com.suman;
import java.util.HashMap;
import java.util.Map;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
import org.apache.log4j.Logger;
/**
* @author Binod Suman
*/
public class TestApplication {
Logger log = Logger.getLogger(TestApplication.class);
public static void main(String[] args) {
TestApplication test = new TestApplication();
test.saveCompany();
}
public void saveCompany(){
log.info("Company data is going to save");
EntityManagerFactory emf;
Map properties = new HashMap();
properties.put("hibernate.connection.driver_class", "com.mysql.jdbc.Driver");
properties.put("hibernate.connection.url", "jdbc:mysql://localhost:3306/sumandb");
properties.put("hibernate.connection.username", "root");
properties.put("hibernate.connection.password", "mysql");
properties.put("hibernate.dialect", "org.hibernate.dialect.MySQLDialect");
properties.put("hibernate.show-sql", "true");
//emf = Persistence.createEntityManagerFactory("jpablogPUnit");
emf = Persistence.createEntityManagerFactory("jpablogPUnit",properties);
EntityManager entityManager = (EntityManager) emf.createEntityManager();
entityManager.getTransaction().begin();
Company company = new Company(120,"TecnoTree","Espoo, Finland");
entityManager.persist(company);
entityManager.getTransaction().commit();
log.info("Company data has been saved");
}
}
来源：https://stackoverflow.com/questions/4106078/dynamic-jpa-connection