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megadix
 Post subject: Detached query with filters on association fields
PostPosted: Sun Jun 03, 2007 12:51 pm 
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Joined: Sun Jun 03, 2007 12:46 pm
Posts: 5
Location: Milan, Italy
Hello,
I have a problem with a DetachedQuery, which seems to be in contrast with Hibernate's documentation.
I'm using Hibernate 3.2 + Spring integration, and my datamodel looks like this:

- class PersonData:
- id, etc...
- contactDetails: ContactDetails

- class ContactDetails:
- firstName, lastName, etc..

I have a PersonDao class with a findByName() method, but it throws an exception when I try to execute the query.

Code:
- CODE START -

public List<PersonData> findByName(String name, SortParameter[] sortParams,
        int first, int count) {

    DetachedCriteria query = DetachedCriteria.forClass(PersonData.class);
    DetachedCriteria contactDetails = query
            .createCriteria("contactDetails");

    SimpleExpression lastNameLike = Restrictions.like("lastName", name,
            MatchMode.START).ignoreCase();
    SimpleExpression firstNameLike = Restrictions.like("firstName", name,
            MatchMode.START).ignoreCase();
    contactDetails.add(Restrictions.disjunction().add(lastNameLike).add(
            firstNameLike));
   
    // adds all the sort parameters
    addSort(query, sortParams);

    return getHibernateTemplate().findByCriteria(query, first, count);
}

- CODE END -


Any ideas ?

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pepelnm
 Post subject:
PostPosted: Sun Jun 03, 2007 2:54 pm 
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Joined: Thu Sep 22, 2005 10:29 am
Posts: 285
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Post the error stack trace.
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megadix
 Post subject:
PostPosted: Sun Jun 03, 2007 6:28 pm 
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Joined: Sun Jun 03, 2007 12:46 pm
Posts: 5
Location: Milan, Italy
pepelnm wrote:
Post the error stack trace.


Oops, you're right, please forgive me :)
Here it is:

Code:
org.springframework.orm.hibernate3.HibernateQueryException: could not resolve property: lastName of: studiodix.model.PersonData; nested exception is org.hibernate.QueryException: could not resolve property: lastName of: studiodix.model.PersonData
Caused by: org.hibernate.QueryException: could not resolve property: lastName of: studiodix.model.PersonData
   at org.hibernate.persister.entity.AbstractPropertyMapping.throwPropertyException(AbstractPropertyMapping.java:43)
   at org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:37)
   at org.hibernate.persister.entity.AbstractEntityPersister.getSubclassPropertyTableNumber(AbstractEntityPersister.java:1327)
   at org.hibernate.persister.entity.BasicEntityPropertyMapping.toColumns(BasicEntityPropertyMapping.java:31)
   at org.hibernate.persister.entity.AbstractEntityPersister.toColumns(AbstractEntityPersister.java:1302)
   at org.hibernate.loader.criteria.CriteriaQueryTranslator.getColumns(CriteriaQueryTranslator.java:434)
   at org.hibernate.loader.criteria.CriteriaQueryTranslator.getColumnsUsingProjection(CriteriaQueryTranslator.java:394)
   at org.hibernate.criterion.SimpleExpression.toSqlString(SimpleExpression.java:45)
   at org.hibernate.criterion.Junction.toSqlString(Junction.java:58)
   at org.hibernate.loader.criteria.CriteriaQueryTranslator.getWhereCondition(CriteriaQueryTranslator.java:334)
   at org.hibernate.loader.criteria.CriteriaJoinWalker.<init>(CriteriaJoinWalker.java:82)
   at org.hibernate.loader.criteria.CriteriaLoader.<init>(CriteriaLoader.java:68)
   at org.hibernate.impl.SessionImpl.list(SessionImpl.java:1550)
   at org.hibernate.impl.CriteriaImpl.list(CriteriaImpl.java:283)
   at org.springframework.orm.hibernate3.HibernateTemplate$35.doInHibernate(HibernateTemplate.java:974)
   at org.springframework.orm.hibernate3.HibernateTemplate.execute(HibernateTemplate.java:362)
   at org.springframework.orm.hibernate3.HibernateTemplate.findByCriteria(HibernateTemplate.java:964)
   at studiodix.dao.PersonDao.findByName(PersonDao.java:42)

[...]


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pepelnm
 Post subject:
PostPosted: Mon Jun 04, 2007 7:31 am 
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What happens if you comment addSort(query, sortParams); line?
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megadix
 Post subject:
PostPosted: Mon Jun 04, 2007 10:16 am 
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Joined: Sun Jun 03, 2007 12:46 pm
Posts: 5
Location: Milan, Italy
Sorry, no change.
addSort() is just a utility method that calls addOrder() if "sortParams" is not null.

Perhaps I've found the reason why it fails: I've mapped ContactDetails as a component...but the documentation says it can be treated as an ordinary association!!
Quote from Hibernate Reference, chapter 8.2:
"Composite elements may appear in queries using the same syntax as associations to other entities."

Uhm, I think that more details are needed. Here are the mapping files:

PersonData.hbm.xml

Code:
<hibernate-mapping package="studiodix.model">
  <class name="PersonData" table="`PERSON_DATA`">
    <id name="id" column="`PERSON_ID`" type="long">
      <generator class="native" />
    </id>
    <component name="contactDetails" class="ContactDetails">
      <property name="title" column="`TITLE`" type="string" length="16" />
      <property name="firstName" column="`FIRST_NAME`" type="string" length="100" />
      <property name="lastName" column="`LAST_NAME`" type="string" length="100" />
      <property name="email" column="`EMAIL`" type="string" length="100" />
      <property name="phone1" column="`PHONE1`" type="string" length="64" />
      <property name="phone2" column="`PHONE2`" type="string" length="64" />
    </component>
  </class>
</hibernate-mapping>



Moreover, if I add "contactDetails." to the restrictions:

Code:
SimpleExpression lastNameLike =
    Restrictions.like("contactDetails.lastName", name, MatchMode.START).ignoreCase();
SimpleExpression firstNameLike =
    Restrictions.like("contactDetails.firstName", name, MatchMode.START).ignoreCase();


I get this absurd error:

Code:
org.springframework.orm.hibernate3.HibernateQueryException: could not resolve property: contactDetails of: studiodix.model.User; nested exception is org.hibernate.QueryException: could not resolve property: contactDetails of: studiodix.model.User
Caused by: org.hibernate.QueryException: could not resolve property: contactDetails of: studiodix.model.User
    at org.hibernate.persister.entity.AbstractPropertyMapping.throwPropertyException(AbstractPropertyMapping.java:43)
    at org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:37)
    at org.hibernate.persister.entity.AbstractEntityPersister.toType(AbstractEntityPersister.java:1310)
    at org.hibernate.loader.criteria.CriteriaQueryTranslator.getPathEntityName(CriteriaQueryTranslator.java:204)
    at org.hibernate.loader.criteria.CriteriaQueryTranslator.createCriteriaEntityNameMap(CriteriaQueryTranslator.java:191)
    at org.hibernate.loader.criteria.CriteriaQueryTranslator.<init>(CriteriaQueryTranslator.java:81)
    at org.hibernate.loader.criteria.CriteriaLoader.<init>(CriteriaLoader.java:59)
    at org.hibernate.impl.SessionImpl.list(SessionImpl.java:1550)
    at org.hibernate.impl.CriteriaImpl.list(CriteriaImpl.java:283)
    at org.springframework.orm.hibernate3.HibernateTemplate$35.doInHibernate(HibernateTemplate.java:974)
    at org.springframework.orm.hibernate3.HibernateTemplate.execute(HibernateTemplate.java:362)
    at org.springframework.orm.hibernate3.HibernateTemplate.findByCriteria(HibernateTemplate.java:964)
    at studiodix.dao.PersonDao.findByName(PersonDao.java:40)


I don't want to fallback to HQL, because I've already put in place a lot of utility code to leverage Detached query mechanism...

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pepelnm
 Post subject:
PostPosted: Mon Jun 04, 2007 1:22 pm 
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Joined: Thu Sep 22, 2005 10:29 am
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I thought contactDetails was an association.

So then, does this work?:
Code:
public List<PersonData> findByName(
    String name, SortParameter[] sortParams, int first, int count
) {

    DetachedCriteria query = DetachedCriteria.forClass(PersonData.class);

    SimpleExpression lastNameLike = Restrictions.like(
        "contactDetails.lastName", name, MatchMode.START
    ).ignoreCase();

    SimpleExpression firstNameLike = Restrictions.like(
        "contactDetails.firstName", name, MatchMode.START
    ).ignoreCase();

    query.add(Restrictions.disjunction().add(lastNameLike).add(firstNameLike));
   
    // adds all the sort parameters
    addSort(query, sortParams);

    return getHibernateTemplate().findByCriteria(query, first, count);
}
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megadix
 Post subject:
PostPosted: Tue Jun 05, 2007 6:10 am 
Newbie

Joined: Sun Jun 03, 2007 12:46 pm
Posts: 5
Location: Milan, Italy
pepelnm wrote:
I thought contactDetails was an association.

So then, does this work?:
Code:
public List<PersonData> findByName(
    String name, SortParameter[] sortParams, int first, int count
) {

    DetachedCriteria query = DetachedCriteria.forClass(PersonData.class);

    SimpleExpression lastNameLike = Restrictions.like(
        "contactDetails.lastName", name, MatchMode.START
    ).ignoreCase();

    SimpleExpression firstNameLike = Restrictions.like(
        "contactDetails.firstName", name, MatchMode.START
    ).ignoreCase();

    query.add(Restrictions.disjunction().add(lastNameLike).add(firstNameLike));
   
    // adds all the sort parameters
    addSort(query, sortParams);

    return getHibernateTemplate().findByCriteria(query, first, count);
}



Unfortunately not. At this point I think that my problem is somewhat related to a deeply hidden error in my mapping files, and that Hibernate just gives back some random error message due to a very unlucky coincidence :(
Oh god I'll have to review ALL my mapping files just to spot this error...
Thank you anyway for your patience, I'll give you credits ;)

Dimitri

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