Iget the following:
13-06-2007 14:52:11:637 ERROR InsertIntoTest main insertion type [org.hibernate.type.LongType@2e1f1f] and selection type [org.hibernate.type.IntegerType@1fd6bea] at position 0 are not compatible [Insert into Population( sessionCacheNumber, itemNumber ) SELECT 1, itemTable.itemNumber FROM ItemTable itemTable]
org.hibernate.QueryException: insertion type [org.hibernate.type.LongType@2e1f1f] and selection type [org.hibernate.type.IntegerType@1fd6bea] at position 0 are not compatible [Insert into Population( sessionCacheNumber, itemNumber ) SELECT 1, itemTable.itemNumber FROM ItemTable itemTable]
at org.hibernate.hql.ast.tree.IntoClause.validateTypes(IntoClause.java:97)
at org.hibernate.hql.ast.tree.InsertStatement.validate(InsertStatement.java:34)
at org.hibernate.hql.ast.HqlSqlWalker.postProcessInsert(HqlSqlWalker.java:603)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.insertStatement(HqlSqlBaseWalker.java:491)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:253)
at org.hibernate.hql.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:228)
at org.hibernate.hql.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:160)
at org.hibernate.hql.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:111)
at org.hibernate.engine.query.HQLQueryPlan.<init>(HQLQueryPlan.java:77)
at org.hibernate.engine.query.HQLQueryPlan.<init>(HQLQueryPlan.java:56)
at org.hibernate.engine.query.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:72)
at org.hibernate.impl.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:133)
at org.hibernate.impl.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:112)
at org.hibernate.impl.SessionImpl.createQuery(SessionImpl.java:1623)
at InsertIntoTest.main(InsertIntoTest.java:31)

code:
Session session =
new AnnotationConfiguration( )
.addAnnotatedClass( Population.class )
.addAnnotatedClass( ItemTable.class )
.configure( "TestHibernate.cfg.xml" )
.buildSessionFactory( ).openSession( );
Transaction transaction = session.beginTransaction( );

session.createQuery(
"Insert into Population( sessionCacheNumber, itemNumber ) " +
"SELECT 1, itemTable.itemNumber FROM ItemTable itemTable" )
.executeUpdate( );


transaction.commit( );
session.close( );

ItemTable.java
@Entity
@Table( name = "item_table" )
public class ItemTable implements java.io.Serializable
{
@Id
@Column( name="item_No", length=65 )
private int itemNumber;

@Column( name="description", length=80 )
private String description;
}

Population.java
@Entity
@IdClass( PopulationPk.class )
@Table( name = "Population" )
public class Population extends PopulationBase
{

@Id
private Long sessionCacheNumber;

@Id
private String itemNumber;
}

PopulationPk.java
@Embeddable
public class PopulationPk extends PopulationBase
{
@Column( name="session_cache_number", precision=18 )
private Long sessionCacheNumber;

@Column( name="item_No", length=65 )
private String itemNumber;
}

PopulationBase.java
public abstract Long getSessionCacheNumber( );

public abstract void setSessionCacheNumber( Long sessionCacheNumber );

public abstract String getItemNumber( );

public abstract void setItemNumber( String itemNumber );

public String toString( )
{
return new ToStringBuilder( this )
.append( "sessionCacheNumber", getSessionCacheNumber( ) )
.append( "itemNumber", getItemNumber( ) )
.toString( );
}

public boolean equals( Object other )
{
if ( !(other instanceof PopulationBase) ) return false;
PopulationBase castOther = (PopulationBase) other;
return new EqualsBuilder( )
.append( getSessionCacheNumber( ), castOther.getSessionCacheNumber( ) )
.append( getItemNumber( ), castOther.getItemNumber( ) )
.isEquals( );
}

public int hashCode( )
{
return new HashCodeBuilder( )
.append( getSessionCacheNumber( ) )
.append( getItemNumber( ) )
.toHashCode( );
}

I need to generate the insert into and dynamically set the sessionCacheNumber. Is there a way to get hibernate to accept it as a long?

Hi,

>>> "SELECT 1, itemTable.itemNumber FROM ItemTable itemTable"
Replace with
"SELECT 1L, itemTable.itemNumber FROM ItemTable itemTable"

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Hi,

I've also an type issue in an insert statement, but mine seems to be harder to solve.

I simplified the example as much as possible. Image the following 2 classes:
Type(int id, ...)
Obj(int id, Type type, ...)

What I want to do is to insert a new Obj with a certain value for "type". In SQL I can write

in case that I know the value of type has to be 1.

In HQL this seems to be much harder. I cannot simply write

without getting a org.hibernate.QueryException: insertion type [org.hibernate.type.ManyToOneType(Type)] and selection type [org.hibernate.type.IntegerType@314af9f7] at position 0 are not compatible [...]

I know that it's possible to cast int to long via "cast(1 as long)". However casting 1 to Type does not work. In this case I get a null pointer exception when I during the creation of the query.

The only workaround I found is to include the Type in the select statement:

But then I assume an unnecessary join is going to be constructed. For a better performance I would like to get rid of this join. Thus, is it somehow possible to work directly with the IDs of persistent objects instead of having to resolve the objects which leads to an unnecessary join?