手动创建EntityManagerFactory?
有谁知道如何手动创建 EntityManagerFactory ?当我说手动时,我的意思是让它消耗一个特殊的 persistence.xml
文件?这是我尝试过的,但一切都让我失败了。
Configuration configuration = new Configuration();
InputStream is = JPAUtil.class.getResourceAsStream("/s-persistence.xml");
configuration.addInputStream(is);
_entityManagerFactory = Persistence.createEntityManagerFactory(
puType.persistenceName, configuration.getProperties());
s-persistence.xml
如下所示:
<?xml version="1.0"?>
<persistence version="1.0" 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_1_0.xsd">
<persistence-unit name="S_DATABASE" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<properties>
<property value="false" name="hibernate.show_sql"/>
<property value="org.hibernate.dialect.OracleDialect" name="hibernate.dialect"/>
<property value="oracle.jdbc.driver.OracleDriver" name="hibernate.connection.driver_class "/>
<property value="jdbc:oracle:thin:@xxx.xxx.xxx.xxx:TEST" name="hibernate.connection.url"/>
<property value="TESTOWNER" name="hibernate.connection.username"/>
<property value="TEST" name="hibernate.connection.password"/>
<property value="2" name="hibernate.connection.isolation"/>
</properties>
</persistence-unit>
</persistence>
运行 java 代码时出现以下错误。
ERROR :: org.hibernate.util.XMLHelper|Error parsing XML: XML InputStream(2) Document is invalid: no grammar found.
ERROR :: org.hibernate.util.XMLHelper|Error parsing XML: XML InputStream(2) Document root element "persistence", must match DOCTYPE root "null".
org.hibernate.MappingException: invalid mapping
这就像它在 persistence.xml
中寻找 hibernate 特定语法。我也尝试过 EJB3Configuration。看起来我的 XML 格式良好,这让我不明白为什么会收到此错误。
Does anyone know how to create an EntityManagerFactory
manually? When I say manually I mean have it consume a special persistence.xml
file? This what I have tried and all has failed me.
Configuration configuration = new Configuration();
InputStream is = JPAUtil.class.getResourceAsStream("/s-persistence.xml");
configuration.addInputStream(is);
_entityManagerFactory = Persistence.createEntityManagerFactory(
puType.persistenceName, configuration.getProperties());
s-persistence.xml
looking like:
<?xml version="1.0"?>
<persistence version="1.0" 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_1_0.xsd">
<persistence-unit name="S_DATABASE" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<properties>
<property value="false" name="hibernate.show_sql"/>
<property value="org.hibernate.dialect.OracleDialect" name="hibernate.dialect"/>
<property value="oracle.jdbc.driver.OracleDriver" name="hibernate.connection.driver_class "/>
<property value="jdbc:oracle:thin:@xxx.xxx.xxx.xxx:TEST" name="hibernate.connection.url"/>
<property value="TESTOWNER" name="hibernate.connection.username"/>
<property value="TEST" name="hibernate.connection.password"/>
<property value="2" name="hibernate.connection.isolation"/>
</properties>
</persistence-unit>
</persistence>
I get the following error when running the java code.
ERROR :: org.hibernate.util.XMLHelper|Error parsing XML: XML InputStream(2) Document is invalid: no grammar found.
ERROR :: org.hibernate.util.XMLHelper|Error parsing XML: XML InputStream(2) Document root element "persistence", must match DOCTYPE root "null".
org.hibernate.MappingException: invalid mapping
It's like it is looking for hibernate specific syntax in the persistence.xml
. I've also tried the EJB3Configuration. It looks like my XML is well formed which is throwing me off as to why I am getting this error.
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我认为您遇到了
可能无法访问的问题,因此无法根据它验证 XML 文件。
尝试不要给出这样的架构位置:
I think you are running into the issue that
may not be reachable, so that validating the XML file against it can not work.
Try to not give the schema location like this: