使用HSQL内存数据库作为JPA数据源 [英] Using HSQL in-memory database as JPA datasource

问题描述

  java.sql.Connection c = DriverManager.getConnection（jdbc： hsqldb：mem：testdb，sa，）; 
 emf = Persistence.createEntityManagerFactory（manager）; 
  

但现在我卡住了。我想将它用作J2SE应用程序中的JPA数据源。我搜遍了整个网页，但所有信息都与J2EE有关。

 < 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 =manager> 
 
< jta-data-source> / *在这里输入什么？* /< / jta-data-source> 
 
<属性> 
 
< property name =hibernate.dialectvalue =org.hibernate.dialect.HSQLDialect/> 
< property name =hibernate.hbm2ddl.autovalue =create-drop/> 
 
< / properties> 
 
< / persistence-unit> 
 
< /持久性> 
  

解决方案

/ *进入这里？* /

好吧，没什么。在Java SE环境中，您必须使用来自JPA提供程序的内置连接池，设置如下所示：

 < persistence xmlns =http://java.sun.com/xml/ns/persistencexmlns：xsi =http://www.w3.org/2001/XMLSchema-instancexsi：schemaLocation = http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsdversion =2.0> 
< persistence-unit name =managertransaction-type =RESOURCE_LOCAL> 
< provider> org.hibernate.ejb.HibernatePersistence< / provider> 
< class> com.acme.Foo< / class> 
 ... 
<属性> 
< property name =javax.persistence.jdbc.drivervalue =org.hsqldb.jdbcDriver/> 
< property name =javax.persistence.jdbc.urlvalue =jdbc：hsqldb：mem：testdb/> 
< property name =javax.persistence.jdbc.uservalue =sa/> 
< property name =javax.persistence.jdbc.passwordvalue =/> 
< property name =hibernate.dialectvalue =org.hibernate.dialect.HSQLDialect/> 
< property name =hibernate.hbm2ddl.autovalue =update/> 
< / properties> 
< / persistence-unit> 
< /余辉> 
  

I have an in-memory data source:

java.sql.Connection c = DriverManager.getConnection("jdbc:hsqldb:mem:testdb", "sa", "");            
emf = Persistence.createEntityManagerFactory("manager");

But now I'm stuck. I want to use it as a JPA data source in a J2SE application. I've scoured the entire web but all info is related to J2EE.

<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="manager">

        <jta-data-source>/*What to enter here?*/</jta-data-source>

        <properties>

            <property name="hibernate.dialect" value="org.hibernate.dialect.HSQLDialect" />
            <property name="hibernate.hbm2ddl.auto" value="create-drop" />

        </properties>

    </persistence-unit>

</persistence>

解决方案

/*What to enter here?*/

Well, nothing. In a Java SE environment, you'll have to use the built-in connection pool from your JPA provider and the setup looks like this:

<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="manager" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <class>com.acme.Foo</class>
    ...
    <properties>
      <property name="javax.persistence.jdbc.driver" value="org.hsqldb.jdbcDriver"/>
      <property name="javax.persistence.jdbc.url" value="jdbc:hsqldb:mem:testdb"/>
      <property name="javax.persistence.jdbc.user" value="sa"/>
      <property name="javax.persistence.jdbc.password" value=""/>
      <property name="hibernate.dialect" value="org.hibernate.dialect.HSQLDialect"/>
      <property name="hibernate.hbm2ddl.auto" value="update"/>
    </properties>
  </persistence-unit>
</persistence>

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