Liberty JavaEE 8 捕获 ConstraintViolationException

Question

Is there any way to catch org.hibernate.exception.ConstraintViolationException in Liberty application server ?

I have a this repository code :

@Stateless
@TransactionManagement(TransactionManagementType.BEAN)
public class MarketRepository {

    @PersistenceContext
    private EntityManager em;

    @Resource
    private UserTransaction tx;

    public void insert(Market market) {
        try {
            tx.begin();
            em.persist(market);
            tx.commit();
        } catch (Exception ignore) {
        }
    }
}     

I always receive this exception :

[INFO] Caused by: org.hibernate.exception.ConstraintViolationException: could not execute statement
[INFO]  at org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:109)
[INFO]  at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:42)
[INFO]  at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:113)
[INFO]  at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:99)
[INFO]  at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.executeUpdate(ResultSetReturnImpl.java:200)
[INFO]  at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:3238)
[INFO]  at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:3763)
[INFO]  at org.hibernate.action.internal.EntityInsertAction.execute(EntityInsertAction.java:107)
[INFO]  at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:604)
[INFO]  at org.hibernate.engine.spi.ActionQueue.lambda$executeActions$1(ActionQueue.java:478)
[INFO]  at java.base/java.util.LinkedHashMap.forEach(LinkedHashMap.java:723)
[INFO]  at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:475)
[INFO]  at org.hibernate.event.internal.AbstractFlushingEventListener.performExecutions(AbstractFlushingEventListener.java:348)
[INFO]  at org.hibernate.event.internal.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:40)
[INFO]  at org.hibernate.event.service.internal.EventListenerGroupImpl.fireEventOnEachListener(EventListenerGroupImpl.java:102)
[INFO]  at org.hibernate.internal.SessionImpl.doFlush(SessionImpl.java:1360)
[INFO]  ... 20 more
[INFO] Caused by: org.postgresql.util.PSQLException: ERROR: duplicate key value violates unique constraint "ukmn9r4yewa9epwp1ekfsbm69gu"
[INFO]   Detail: Key (exchange, symbol)=(COINEX, ZECUSDT) already exists.
[INFO]  at org.postgresql.core.v3.QueryExecutorImpl.receiveErrorResponse(QueryExecutorImpl.java:2533)
[INFO]  at org.postgresql.core.v3.QueryExecutorImpl.processResults(QueryExecutorImpl.java:2268)
[INFO]  at org.postgresql.core.v3.QueryExecutorImpl.execute(QueryExecutorImpl.java:313)
[INFO]  at org.postgresql.jdbc.PgStatement.executeInternal(PgStatement.java:448)
[INFO]  at org.postgresql.jdbc.PgStatement.execute(PgStatement.java:369)
[INFO]  at org.postgresql.jdbc.PgPreparedStatement.executeWithFlags(PgPreparedStatement.java:159)
[INFO]  at org.postgresql.jdbc.PgPreparedStatement.executeUpdate(PgPreparedStatement.java:125)
[INFO]  at jdk.internal.reflect.GeneratedMethodAccessor1090.invoke(Unknown Source)
[INFO]  at java.base/java.lang.reflect.Method.invoke(Method.java:567)
[INFO]  at org.postgresql.ds.PGPooledConnection$StatementHandler.invoke(PGPooledConnection.java:428)
[INFO]  at jdk.proxy11/jdk.proxy11.$Proxy136.executeUpdate(Unknown Source)
[INFO]  at com.ibm.ws.rsadapter.jdbc.WSJdbcPreparedStatement.executeUpdate(WSJdbcPreparedStatement.java:520)
[INFO]  at [internal classes]
[INFO]  at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.executeUpdate(ResultSetReturnImpl.java:197)
[INFO]  ... 31 more

this is my server.xml config :

 <dataSource jndiName="jdbc/jta-datasource" transactional="true">
        <jdbcDriver id="database-driver" libraryRef="project-libs"/>
        <properties databaseName="${database.name}" serverName="${database.hostname}" portNumber="${database.port}"
                    user="${database.username}" password="${database.password}"/>
    </dataSource>    

so, I want to ignore this exception, but still receive that !

How can fix this problem ?

UPDATE:
persistence.xml

<persistence 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"
             version="2.1">
    <persistence-unit name="db-connection" transaction-type="JTA">
        <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
        <jta-data-source>jdbc/jta-datasource</jta-data-source>
        <properties>
            <property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQL10Dialect"/>
        </properties>
    </persistence-unit>
</persistence>    

MarketJob:

import javax.ejb.*;

@Singleton
public class MarketJob {

    @EJB
    private MarketRepository repo;

    @Lock(LockType.READ)
    @Schedule(second = "59", minute = "59", hour = "*", persistent = false)
    private void execute() {
        List<Market> marketList = // read data from external api;
        for (Market market : marketList) {
            repo.insert(market);
        }
    }
}

Answer 1

If you use em.merge(market); instead of em.persist(market); JPA translates that to an insert-if-absent-else-update, which will avoid violating the constraint in your database that every (exchange, symbol) pairing be unique.

This may not work if the unique constraint isn't a primary key. Or even if it is ... My experience is you just can't catch an exception that the persistence container is throwing. One solution would be to make sure there isn't already a record with the same exchange and symbol:

String jpql = "select count(*) from Market where m.exchange = ? and m.symbol = ?";
tx.begin();
int count = em.createQuery(jpql, Integer.class)
.setParameter(1, market.getExchange())
.setParameter(2, market.getSymbol()).getSingleResult();
if(count == 0){
  em.persist(market);
  tx.commit();
}

Answer 2

You just need to put arround the repo.insert(market) a try/catch block that catches SQLIntegrityConstraintViolationException , and do whatever you want in the catch block.

Off-topic recomendation for your code

UserTransaction: At least in the example you are using you don't need that. All Session Bean (aka EJB) (that is Stateless, Stateful and Singleton) provide transactionality by default. That means that before any business method a new transaction will automatically start, and will commit it at the end of the method. If the method throws an Exception the transaction will automatically Rollback.

If the business method is called from another EJB it will reuse the same transaction.

That is the default behaviour, and is really really powerful, allowing to manage most of the business cases without manually managing transactions.