How can I filter data on email property?
Hibernate version: 3

Mapping documents:
<class name="package.User" table="user">
<id name="id" column="id" type="long">
<generator class="sequence">
<param name="sequence">SEQ</param>
</generator>
</id>

<property name="login" type="string" column="login" />

<join table="info">
<key column="user_id" />
<property name="info" type="package.InfoUserType">
<column name="email" />
</property>
</join>
</class>

Code between sessionFactory.openSession() and session.close():
String HQL = "from package.User user where user.info.email like :email";
Query query = getSession().createQuery(HQL);
List<Site> results = query.list();

Full stack trace of any exception that occurs:
org.hibernate.QueryException: could not resolve property: email of: package.User [from package.User user where user.info.email like :email]
at org.hibernate.persister.entity.AbstractPropertyMapping.propertyException(AbstractPropertyMapping.java:44)
at org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:38)
at org.hibernate.persister.entity.AbstractEntityPersister.toType(AbstractEntityPersister.java:1354)
at org.hibernate.hql.ast.tree.FromElementType.getPropertyType(FromElementType.java:279)
at org.hibernate.hql.ast.tree.FromElement.getPropertyType(FromElement.java:386)
at org.hibernate.hql.ast.tree.DotNode.getDataType(DotNode.java:553)
at org.hibernate.hql.ast.tree.DotNode.prepareLhs(DotNode.java:227)
at org.hibernate.hql.ast.tree.DotNode.resolve(DotNode.java:178)
at org.hibernate.hql.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:94)
at org.hibernate.hql.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:90)
at org.hibernate.hql.ast.HqlSqlWalker.resolve(HqlSqlWalker.java:728)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.expr(HqlSqlBaseWalker.java:1216)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.exprOrSubquery(HqlSqlBaseWalker.java:4041)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.comparisonExpr(HqlSqlBaseWalker.java:3669)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.logicalExpr(HqlSqlBaseWalker.java:1762)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.whereClause(HqlSqlBaseWalker.java:776)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:577)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:281)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:229)
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)

Name and version of the database you are using:
Oracle 9i
The generated SQL (show_sql=true):
NA
Debug level Hibernate log excerpt:

Problems with Session and transaction handling?

you should add the join table to from in hql.
like :
from a inner join b where...

in your case, email field cannot be found, since the info class is not added.

Thanks for the help.

I have a class User something like this

class User {
private Info info;

public Info getInfo(){
return this.info;
}

public void setInfo(Info info) {
this.info = info;
}
}

and in the same way Info has email property, why that property is not visible?
we should be able to filter on email, because hibernate is already qurieing the info table in DB.