java.lang.ClassCastException: java.math.BigInteger cannot be cast to java.lang.Long

Question

I want to return number of rows using native sql. But console says me java.math.BigInteger cannot be cast to java.lang.Long. What's wrong? This is my method:

public Long getNumRows(Integer id){

        Session session = null;

        session = this.sessionFactory.getCurrentSession();
        Query query = session
                .createSQLQuery("SELECT COUNT(*) FROM controllnews WHERE news_id="
                        + id + ";");
        List firstResult = query.list();

        return (Long) firstResult.get(0);


    }

Rohit Jain · Accepted Answer · 2013-08-13 20:27:23Z

12

Use BigInteger#longValue() method, instead of casting it to Long:

return firstResult.get(0).longValue();

Seems like firstResult.get(0) returns an Object. One option is to typecast to BigInteger:

return ((BigInteger)firstResult.get(0)).longValue();

But don't do this. Instead use the way provided by Nambari in comments. I'm no user of Hibernate, so I can't provide Hibernate specific solution.

edited Aug 13, 2013 at 20:27

answered Aug 13, 2013 at 20:20

Rohit Jain

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7 Comments

kosa Over a year ago

(OR) use hibernate scalars docs.jboss.org/hibernate/orm/3.6/reference/en-US/html/…

Rohit Jain Over a year ago

@Nambari. Oh! I've not used Hibernate though.

John Smith Over a year ago

The method longValue() is undefined for the type Object

kosa Over a year ago

And your answer will not work because return type of firstResult.get(0) is Object.

Rohit Jain Over a year ago

@JohnSmith. Although it works, please look into Hibernate Scalars. That's perhaps the ideal way here.

|

Community · Accepted Answer · 2017-05-23 12:10:04Z

I faced same problem, I used Hibernate Scalar queries as suggested in the comment of @Rohit Jain's answer. Thanks @nambari for comment.

Coming to the problem we have,

Query query = session
            .createSQLQuery("SELECT COUNT(*) FROM controllnews WHERE news_id="
                    + id + ";");

These will return a List of Object arrays (Object[]) with scalar values for each column in the controllnews table. Hibernate will use ResultSetMetadata to deduce the actual order and types of the returned scalar values.

To avoid the overhead of using ResultSetMetadata, or simply to be more explicit in what is returned, one can use addScalar():

Query query = session
            .createSQLQuery("SELECT COUNT(*) as count FROM controllnews WHERE news_id="
                    + id + ";").addScalar("count", LongType.INSTANCE);

This will return Object arrays, but now it will not use ResultSetMetadata but will instead explicitly get the count column as Long from the underlying resultset.

How the java.sql.Types returned from ResultSetMetaData is mapped to Hibernate types is controlled by the Dialect. If a specific type is not mapped, or does not result in the expected type, it is possible to customize it via calls to registerHibernateType in the Dialect.

You can use Query#setParameter method to avoid any mistakes in query as

Query query = session
           .createSQLQuery("SELECT COUNT(*) as count FROM controllnews WHERE news_id= :id")
           .addScalar("count", LongType.INSTANCE).setParameter("id",id);

One confusion when you refer Scalar queries docs 4.0, addScalar method has second parameter Hibernate.LONG, remember it has been deprecated since Hibernate version 3.6.X
Here is the deprecated document, so you have to use LongType.INSTANCE

Related link

Hibernate javadocs

Ajinz · Accepted Answer · 2019-08-14 09:43:14Z

4

  String query ="select count(*) from tablename";       
  Number count = (Number) sessionFactory.getCurrentSession().createSQLQuery(query).uniqueResult();
  Long value = count.longValue();

I tried this it works well

answered Aug 14, 2019 at 9:43

Ajinz

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