Hibernate Community • View topic - Hibernate and SQLServer uniqueidentifier problem

View unanswered posts | View active topics

Board index » Hibernate & Java Persistence » Hibernate Users

All times are UTC - 5 hours [ DST ]

Hibernate and SQLServer uniqueidentifier problem

Page 1 of 1

[ 5 posts ]

Previous topic | Next topic

Author

Message

gaitonde

Post subject: Hibernate and SQLServer uniqueidentifier problem

Posted: Thu Nov 27, 2003 12:58 pm

Newbie

Joined: Mon Oct 06, 2003 10:33 am
Posts: 2

Hello:

Sorry if this topic has been hashed through already, but I haven't been able to find a conclusive answer.

Has anyone gotten Hibernate to work with the SQL Server "uniqueidentifier" column as the primary key? That is, allowing SQL Server to generate the id for you (via the SQL Server "newid"). The problem surfaces when multiple objects are persisted sequentially using the same session. I've tried different generators: native, uuid.hex, assigned.

I believe the problem boils down to the fact that you cannot get the newly generated id once a row is inserted. In effect, "select @@identity" does not work after a new row has been inserted if the primary key is of type uniqueidentifier. Resulting from this is that Hibernate cannot cache the object properly. I receive a "Object has already been loaded error".

Any help on how to get Hibernate and SQL Server working with "uniqueidentifier" would be greatly appreciated.

Thanks in advance.

--Salil

Top

andreas_eriksson

Post subject:

Posted: Fri Nov 28, 2003 10:43 am

Regular

Joined: Wed Aug 27, 2003 2:55 am
Posts: 90
Location: Sweden

Hi,

Subclass Hibernate UUIDGenerator and make your own GUID's.

The MS SQL Server GUID format is XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX (8-4-4-4-12).

DO NOT generate the brackets shown in MS SQL Server, there ONLY MS compatible (I think).

An alternative to making your own GUID could be done by something like this:

1. Create a GUID table with on column as UID and the other as whatever.
2. In your UUIDGenerator method generate() insert a value to the 'whatever' column in table GUID so a UID is generated.
3. Read the generated UID from table and return as result in method generate()

Kind regards, Andreas

Top

gaitonde

Post subject:

Posted: Fri Nov 28, 2003 3:49 pm

Newbie

Joined: Mon Oct 06, 2003 10:33 am
Posts: 2

andreas_eriksson wrote:

Hi,

Subclass Hibernate UUIDGenerator and make your own GUID's.

The MS SQL Server GUID format is XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX (8-4-4-4-12).

But is this approach any different from me using the "assigned" generator and providing my own guid? In other words, using the subclassing approach, won't Hibernate still try to get the id after row insertion, and result in the same problem?

Thanks,
Salil

Top

andreas_eriksson

Post subject:

Posted: Fri Nov 28, 2003 8:38 pm

Regular

Joined: Wed Aug 27, 2003 2:55 am
Posts: 90
Location: Sweden

Since the GUID is generated in the app. there's no need for Hibernate performing the 'SELECT @@IDENTITY'.

The uniqueidentifier type in MS SQL Server could still be used, but NOT generated in the db.
The primary key should NOT be defined as type varchar even though we generate the GUIDs strings in the app. The uniqueidentifier type could serve as "validation" in the db. 'Cause if you try to insert a String that's not following the format 8-4-4-4-12 (HEX String), the db will protest.

Regards, Andreas

Top

dada

Post subject: Typ problems with ID of MS SQL 2000

Posted: Mon Feb 27, 2006 8:42 am

Newbie

Joined: Mon Feb 27, 2006 8:25 am
Posts: 3

Hi

I am new to hibernate so my question is probably not that interesting.

My table in the DB uses a own column with automatically assigned GUIDS.
I tried following but it doesn't work.

Thank you very much

the greenhorn

Code:

Initial SessionFactory creation failed.org.hibernate.MappingException: Could not determine type for: guid, for columns: [org.hibernate.mapping.Column(GUIDModulID)]
Exception in thread "main" java.lang.ExceptionInInitializerError
   at util.HibernateUtil.<clinit>(HibernateUtil.java:17)
   at verwaltung.Modulmanager.createModul(Modulmanager.java:32)
   at verwaltung.Modulmanager.main(Modulmanager.java:17)
Caused by: org.hibernate.MappingException: Could not determine type for: guid, for columns: [org.hibernate.mapping.Column(GUIDModulID)]
   at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:266)
   at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:253)
   at org.hibernate.mapping.RootClass.validate(RootClass.java:193)
   at org.hibernate.cfg.Configuration.validate(Configuration.java:982)
   at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1146)
   at util.HibernateUtil.<clinit>(HibernateUtil.java:13)
   ... 2 more

Code:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd" >
<hibernate-mapping>
   <class name="verwaltung.Modul" table="Modul">
      <id name="id" column="GUIDModulID" >
         <generator class="guid"></generator>
      </id>
      <property name="Name"> <column name="ModulBezeichnung"></column></property>
   </class>
</hibernate-mapping>

Code:

package verwaltung;
import org.hibernate.id.GUIDGenerator;

public class Modul {

   private GUIDGenerator id;
   
   private String fName;
   
   public Modul(){}

   public String getName() {
      return fName;
   }

   public void setName(String name) {
      fName = name;
   }

   public GUIDGenerator getId() {
      return id;
   }

   private void setId(GUIDGenerator id) {
      this.id = id;
   };
   
}

Top

Page 1 of 1

[ 5 posts ]

Board index » Hibernate & Java Persistence » Hibernate Users

All times are UTC - 5 hours [ DST ]

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum