Develop Reference

Should I instantiate EntityManager for every method in repository [duplicate]

I am not using Spring so I am creating an instance of EntityManager within a class.
I used Hibernate-Eclipse reverse engineering to auto-generate the classes. These classes all has an instance of EntityManager.
I am not 100% sure how Hibernate works with the EntityManager so I am wondering if it is okay that so many instances of this class (EntityManager) are made, for example, will there be problems with transactions?
Should I just make a separate class that distributes a static instance of an EntityManager for all my other classes? or does it not matter?
EDIT: I see there's something called #PersistenceContext, it doesn't seem to load my persistence.xml as a bean in to the instance variable, does this feature require spring? (I get null pointer exception, because it was never injected)
snip of code from where I attempt to use #persistencecontext
#PersistenceContext(unitName = "manager1")
private EntityManager entityManager;
my persistence.xml
<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="manager1" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<properties>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
<property name="javax.persistence.jdbc.user" value="root"/>
<property name="javax.persistence.jdbc.password" value="mypassword"/>
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost/ptbrowserdb"/>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect"/>
</properties>
</persistence-unit>
</persistence>

See this Article: JPA Architecture it explain it very well.
In General you need a single Entity Manager per transaction. And this Entity Manager must not be used in two transactions at the same time.
Clairification: I mean, do not use a single Entity Manager for different unit of works. Typical one transaction in one unit of work, if you have different transactions of one unit of work, then you can use the same Entity Manager
If you use Spring then Spring do this handling for you if you use the #PersistenceContext annotation to inject the EntityManager. Per default Spring "bind" the the injected EntityManager (via a proxy) to the current transaction. (And the transaction is "bound" to the thread.)
#See Spring Reference 13.5.2 Implementing DAOs based on plain JPA - it contains a interesting paragagraph after the code examples.

You need a dependency injection framework like Spring or Google Guice to inject objects into your class otherwise it may not be injected automatically for you.
Basically this is an annotation provided by JPA which will work with in tandem with hibernate or any other ORM framework per say but you need a DI framework to inject the objects.
Regarding the single instance of entity manager i don't think you need that if you go by Spring since it takes care of managing the instances and the transactions for you by tying your entity manager with the jpa transaction.

Dynamically create EntityManager / Connect to custom host/DB every time in Hibernate

I have an application running on Websphere Liberty which should compare tables from 2 databases/schemas.
The user should be able to input the connection data, like the host and the credentials.
I'm using Hibernate to access the application DB.
I've tried to use multiple Persistence Units, one for the application DB and one for all other DBs.
But i got 2 problems:
i get an "Illegal attempt to enlist multiple 1PC XAResources"
error sometimes
can query the 2 DBs with
user-submitted credentials, but i get no results except if i connect
to the same db listed in server.xml file as DataSource
This are the DataSources on the server.xml on the server (dbs are oracle dbs)
<dataSource id="MyAppDS" jndiName="jdbc/MyDS" type="javax.sql.ConnectionPoolDataSource">
<jdbcDriver javax.sql.ConnectionPoolDataSource="oracle.jdbc.pool.OracleConnectionPoolDataSource" libraryRef="OracleSQLLib"/>
<connectionManager agedTimeout="30m" connectionTimeout="10s" maxPoolSize="20" minPoolSize="5"/>
<properties password="..." url="jdbc:oracle:thin:#...:1521:..." user="..."/>
</dataSource>
<dataSource id="OtherOracle" jndiName="jdbc/OtherOracle" type="javax.sql.ConnectionPoolDataSource">
<jdbcDriver javax.sql.ConnectionPoolDataSource="oracle.jdbc.pool.OracleConnectionPoolDataSource" libraryRef="OracleSQLLib"/>
<connectionManager agedTimeout="30m" connectionTimeout="10s" maxPoolSize="20" minPoolSize="5"/>
<properties password="..." url="jdbc:oracle:thin:#127.0.0.1:1521:XE" user="..."/>
</dataSource>
This is persistence.xml on the EJB module
<?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="main-persistence">
<jta-data-source>jdbc/MyDS</jta-data-source>
<class>classes...</class>
<properties>
<property name="hibernate.transaction.jta.platform"
value="org.hibernate.service.jta.platform.internal.WebSphereExtendedJtaPlatform" />
<property name="hibernate.dialect"
value="org.hibernate.dialect.Oracle9iDialect" />
<property name="hibernate.temp.use_jdbc_metadata_defaults"
value="false" />
</properties>
</persistence-unit>
<persistence-unit name="other-persistence" transaction-type="RESOURCE_LOCAL">
<non-jta-data-source>jdbc/OtherOracle</non-jta-data-source>
<class>classes...</class>
<properties>
<property name="hibernate.transaction.jta.platform"
value="org.hibernate.service.jta.platform.internal.WebSphereExtendedJtaPlatform" />
<property name="hibernate.dialect"
value="org.hibernate.dialect.Oracle9iDialect" />
<property name="hibernate.temp.use_jdbc_metadata_defaults"
value="false" />
</properties>
</persistence-unit>
On the Java Bean i use the EntityManagerFactory
#PersistenceUnit(unitName = "other-persistence")
private EntityManagerFactory emf;
And i create the entity manager with custom credentials like this
Map<String, String> properties = new HashMap<String, String>();
properties.put("hibernate.connection.driver_class", "oracle.jdbc.OracleDriver");
properties.put("hibernate.connection.url", myCustomCreatedConnectionUrl);
properties.put("hibernate.connection.username", customUser);
properties.put("hibernate.connection.password", customPassword);
properties.put("hibernate.dialect", "org.hibernate.dialect.Oracle9iDialect");
properties.put("hibernate.show-sql", "true");
EntityManager entityManager = emf.createEntityManager(properties);
If i check EntityManager properties with getProperties everything seems to be right. But the queries works only if the credentials/host are = to the DataSource. Otherwise i get no results (but no errors)
What could the problem be?
Is there a way to use just one Persistence Unit but with custom host/credentials for different queries?

Regarding the first error, ""Illegal attempt to enlist multiple 1PC XAResources", this occurs because you are using both resources within the same transaction. I can see <non-jta-data-source>jdbc/OtherOracle</non-jta-data-source> in your configuration, which indicates that you might have intended for jdbc/OtherOracle to be a non-enlisting resource. To make that work, the data source itself needs to be configured as non-enlisting. You can do this with the transactional="false" attribute as follows:
<dataSource id="OtherOracle" jndiName="jdbc/OtherOracle" type="javax.sql.ConnectionPoolDataSource" transactional="false">
...
On the other hand, if you actually want both resources to enlist in the transaction, then you need to use XADataSource rather than ConnectionPoolDataSource. Here is an example of how to do that (notice that both type type under dataSource and the attribute & class under jdbcDriver must be updated for this:
<dataSource id="MyAppDS" jndiName="jdbc/MyDS" type="javax.sql.XADataSource">
<jdbcDriver javax.sql.XADataSource="oracle.jdbc.xa.client.OracleXADataSource" libraryRef="OracleSQLLib"/>
<connectionManager agedTimeout="30m" connectionTimeout="10s" maxPoolSize="20" minPoolSize="5"/>
<properties password="..." url="jdbc:oracle:thin:#...:1521:..." user="..."/>
</dataSource>
<dataSource id="OtherOracle" jndiName="jdbc/OtherOracle" type="javax.sql.XADataSource">
<jdbcDriver javax.sql.XADataSource="oracle.jdbc.xa.client.OracleXADataSource" libraryRef="OracleSQLLib"/>
<connectionManager agedTimeout="30m" connectionTimeout="10s" maxPoolSize="20" minPoolSize="5"/>
<properties password="..." url="jdbc:oracle:thin:#127.0.0.1:1521:XE" user="..."/>
</dataSource>
In the second question, I think you are saying the different users cannot see the data. Could this be because the different database users are using different schema and do not have access each others' data? If you can get all of the users using a common schema, then adding #Table(schema="YOUR_SCHEMA_NAME") to the JPA #Entity might help. JavaDoc for the Table annotation can be found here.

I have recently had the same request where a user can connect to application and enter their database details. From this point on every JPA Query is executed on this new connection.
The additional problem was that there were more than 100 bases to choose from and creating all datasources at start was not a good idea.
We have thus created a single RoutingDatasource that manages the JDBC connections. We were heavily inspired by Spring's implementation org.springframework.jdbc.datasource.lookup.AbstractRoutingDataSource.
The main idea is:
You extend AbstractDataSource and getConnection() method which is called on every JPA query. You are basically tinkering the JDBC layer under the JPA layer.
You then create a new datasource according to what your user entered
Set Your routing datasource as main datasource for your entity manager
Optionally add some caching to avoid recreating datasources every time (or implement some pool)
Here is a demo class:
public class RoutingDataSource extends AbstractDataSource {
...
...
#Override
public Connection getConnection() throws SQLException {
return determineTargetDataSource().getConnection();
}
DataSource determineTargetDataSource() {
EmployeeDatabase lookupKey = determineCurrentLookupKey();
DataSource dataSource = this.datasources.get(lookupKey);
if (dataSource == null) {
logger.debug("Datasource not found. Creating new one");
SQLServerDataSource newDatasource = new SQLServerDataSource();
newDatasource.setURL("jdbc:sqlserver://" + lookupKey.getDatabaseHost());
newDatasource.setPassword(dbPass);
datasources.put(lookupKey, newDatasource);
dataSource = newDatasource;
} else {
logger.debug("Found existing database for key " + lookupKey);
}
logger.debug("Connecting to " + dataSource);
return dataSource;
}
}

Hibernate configuration with MongoDB

I've been trying to mess around with MongoDB and Hibernate in Java. I'm having some troubles with the configuration file for it.
I've already used Hibernate in the past with SQL DB, but it seems that the config file has to be quite different for MongoDB.
According to this documentation, it should looks something like this:
<?xml version="1.0" encoding="UTF-8"?>
<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="eshop" transaction-type="JTA">
<provider>org.hibernate.ogm.jpa.HibernateOgmPersistence</provider>
<class>org.hsnr.rest.domain.entities.Address</class>
<class>org.hsnr.rest.domain.entities.Order</class>
<class>org.hsnr.rest.domain.entities.Person</class>
<class>org.hsnr.rest.domain.entities.Product</class>
<class>org.hsnr.rest.domain.entities.User</class>
<properties>
<property name="hibernate.transaction.jta.platform"
value="org.hibernate.service.jta.platform.internal.JBossStandAloneJtaPlatform" />
<property name="hibernate.ogm.datastore.provider"
value="mongodb" />
</properties>
</persistence-unit>
</persistence>
However, when I try to create a session with
new Configuration().configure().buildSessionFactory();
I get a following error:
org.hibernate.internal.util.config.ConfigurationException: Unable to perform unmarshalling at line number 5 and column 28 in RESOURCE hibernate.cfg.xml. Message: cvc-elt.1: Cannot find the declaration of element 'persistence'.
Is my aproach wrong or is there something I overlooked?

You can try basic test set up like below for your config.
EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory( "eshop" );
EntityManager entityManager = entityManagerFactory.createEntityManager();
entityManager.getTransaction().begin();
// perform operations here
entityManager.getTransaction().commit();
entityManager.close();
entityManagerFactory.close();

From the javadoc of configure():
Use the mappings and properties specified in an application resource
named hibernate.cfg.xml.
You are setting the persistence.xml instead. Using using javax.persistence.Persistence should work:
EntityManagerFactory emf = Persistence.createEntityManagerFactory( "eshop" );
If for some reason you need the session factory instead, but you are working with JPA, you can obtain it using the unwrap() method
SessionFactory sf = emf.unwrap( SessionFactory.class );
UPDATE:
You can also create the factory programmatically, there is a class OgmConfiguration (that extends Configuraiton):
OgmConfiguration configuration = new OgmConfiguration();
// This is optional, if you want to set some options using
// a fluent API
configuration.configureOptionsFor( MongoDB.class )
.writeConcern( WriteConcernType.UNACKNOWLEDGED );
SessionFactory sf = configuration
.addAnnotatedClass( org.hsnr.rest.domain.entities.Address.class )
// ... Other annotated classes
.setProperty( OgmProperties.DATABASE, "mongodb_database" )
.setProperty( OgmProperties.DATASTORE_PROVIDER, DatastoreProviderType.MONGODB.name() )
// All this properties are optional, appropriate default will be used if missing
.setProperty( OgmProperties.CREATE_DATABASE, "false" )
.setProperty( OgmProperties.USERNAME, "username" )
.setProperty( OgmProperties.PASSWORD, "password" )
.setProperty( OgmProperties.HOST, "localhost:12897" )
// Check MongoDBProperties for MongoDB specific options
.setProperty( MongoDBProperties.AUTHENTICATION_MECHANISM, AuthenticationMechanismType.BEST.name() )
.buildSessionFactory();
In this example I've added several options to give an overview but if you are using defaults and you don't need the authentication, only the database name and the datastore provider are needed

use bellow code :
<hibernate-configuration
xmlns="http://www.hibernate.org/xsd/hibernate-configuration"
xsi:schemaLocation="http://www.hibernate.org/xsd/hibernate-configuration hibernate-configuration-4.0.xsd"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">

Container-managed EntityManager: multiple managers for single persistence unit?

I have a use-case where I think I need two entity managers, that access the same persistence unit. So essentially I want two persistence contexts on the same database. Is this possible via #PersistenceContext annotations?
I want to write something like the following, but don't know how to tell JPA to inject two different manager instances.
#PersistenceContext(type = PersistenceContextType.EXTENDED)
private EntityManager entityManager;
#PersistenceContext(type = PersistenceContextType.EXTENDED)
private EntityManager otherEntityManager;
I think I could switch to application-managed transactions, then I could just create another one using a factory. But I don't want to manage the transactions by myself, if it's not absolutely necessary.

There is some ambiguity in your statement. Are you constrained by using only one 'Persistent Unit'? It is not same as Constrained by using Single Datasource.
You can create multiple persistent units for a single datasource. So, if you are not constrained by the number of persistent units you can create, You can in persistence.xml declare 2 persistent units for the same datasource like below
<?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="PU1"
transaction-type="JTA">
<jta-data-source>jdbc/myDS</jta-data-source>
<!-- Other properties -->
</persistence-unit>
<persistence-unit name="PU2"
transaction-type="JTA">
<jta-data-source>jdbc/myDS</jta-data-source>
<!-- Other properties -->
</persistence-unit>
</persistence>
Then, you can create 2 entitymanagers like below
#PersistenceContext(unitName="PU1", type = PersistenceContextType.EXTENDED)
private EntityManager entityManager;
#PersistenceContext(unitName="PU2", type = PersistenceContextType.EXTENDED)
private EntityManager otherEntityManager;
Hope this helps.

JPA deceleration exception

I am using jboss5.1.x EJB3.0
I am trying my first time with JPA, and I get this exception when I run the server:
java.lang.IllegalArgumentException: Can't find a persistence unit named 'java:/mracDS'
..
this is my "DAO" entity which is responsible on all the JPA entities:
#Stateless
public class ECMSEntityManagerDao implements ECMSEntityManagerDaoLocal, ECMSEntityManagerDaoRemote
{
#PersistenceContext(unitName = "java:/mracDS")
EntityManager em;
public ArrayList<T01CounterCalls> getClocksDetailsFromVantive() throws SQLException
{
return (ArrayList<T01CounterCalls>) em.createQuery ("from T01CounterCalls as data").getResultList ();
}
}
I looked a bit in the net:
I never declared persistence.xml
and if I do, what should be declared inside?
thanks,
ray.

From the look of it my guess is that you are confusing a persistence unit with a data source.
These two may feel similar, but they aren't. Very simply said a persistence unit is a set of classes plus an associated data source. In the most basic form, a persistence unit merely couples to a data source:
persistence.xml:
<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence" version="1.0"
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="mracPU">
<jta-data-source>java:/mracDS</jta-data-source>
</<persistence-unit>
</persistence>
Then use the persistence unit name with the injection annotations:
#Stateless
public class ECMSEntityManagerDao implements ECMSEntityManagerDaoLocal, ECMSEntityManagerDaoRemote {
#PersistenceContext(unitName = "mracPU")
EntityManager em;
}