Hibernate Query Cache

Hibernate provides sophisticated caching mechanisms to improve the performance. Apart from
First-level, Second-Level and query cache, Hibernate can be configured to make use of some open source cache solutions like EHCache or JBoss Cache etc.

If your Hibernate query yields huge but similar results every time it is executed, it is a good idea to cache the results in hibernate to improve the performance. This could be achieved by first enable the hibernate query cache in the hibernate configuration
in properties file

hibernate.cache.use_query_cache true

or xml configuration file like

<property name="hibernate.cache.use_query_cache">true</property>

and while executing the query specify the cacheability like this

sess.createQuery("from.......")
    .setCacheable(true)
    .list();

SQLGrammarException could not fetch initial value for increment generator

In the earlier versions of Hibernate, may be hibernate 2.x, if you use increment generator on a new table then you might get this error. If you really need to use the increment generator on the new table then you need to create a dummy row on that table.
This issue is not present in the current version of hibernate i.e. Hibernate 3.x

HibernateSystemException- Don’t change the reference to a collection with cascade=”all-delete-orphan”

This is not an error which I got. This is an error that bugged one of my friend for a few days.

The error is

org.springframework.orm.hibernate3.HibernateSystemException: Don’t change the reference to a collection with cascade=”all-delete-orphan”: com.test.Cat.kittens; nested exception is org.hibernate.HibernateException: Don’t change the reference to a collection with cascade=”all-delete-orphan”: com.test.Cat.kittens on 10.01.2009@15.26.33

<hibernate-mapping package="com.test">
       <class name="Cat" lazy="false">
			   ......
			   ......
			   ......

               <set name="kittens" table="kitten" order-by="kitten_name asc" inverse="true"                             cascade="all-delete-orphan">
               <key column="cat_id"/>
               <one-to-many class="Kitten"/>
           </set>
       </class>
</hibernate-mapping>

We researched a few solutions on the net. The following solutions are based mainly on the two sources
http://forum.springframework.org/showthread.php?t=25809
http://www.hibernate.org/117.html#A3

Since I myself did not get this problem I can’t give much details. And the solutions below may or may not work for you.

There are two ways to solve this problem
i) If posible switch to the latest version of the Hibernate. Even though it is not a bug in the Hibernate, some people reported that somehow this problem doesn’t appear in the new version of hibernate.

Caution: If your application uses Hibernate extensively then you might consider testing the application throughly after upgrade. If you can’t upgrade for some reasons, then read on

ii) If you can’t upgrade then as per the solutions provided on the forums, you should call clear() if you need to remove the collections marked as all-delete-orphan.
Fair enough. Makes sense, since you are cascading the delete you should inform Hibernate if you intend to remove the collection.

For example,
Instead of doing something like this.

	public void removeKittens(){
	  //Load Cat with kittens
	  Cat cat = getHibernateTemplate().get(Cat.class, 1234);
	  cat.setKittens(null);
	  getHibernateTemplate().save(cat);
	}

Do something like this

	public void removeKittens(){
	  //Load Cat with kittens
	  Cat cat = getHibernateTemplate().get(Cat.class, 1234);
	  cat.getKittens.clear();
	  getHibernateTemplate().save(cat);
	}

As I said this is not a bug in Hibernate’s older version. But somehow in the new version, as people reported on the forums, this problem is not there. That means you are not required to call clear(). I am not sure because I have not faced this problem myself.

Hibernate error : org.hibernate.hql.ast.QuerySyntaxException: unexpected end of subtree

I got this error today.

Caused by: org.hibernate.hql.ast.QuerySyntaxException: unexpected end of subtree [select count(*)  from com.test.Fruits as fruit where fruit.fruitId in () order by fruit.fruitName desc ]
	at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:31)
	at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:24)
	at org.hibernate.hql.ast.ErrorCounter.throwQueryException(ErrorCounter.java:59)
	at org.hibernate.hql.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:235)
	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.SessionImpl.list(SessionImpl.java:1113)
	at org.hibernate.impl.QueryImpl.list(QueryImpl.java:79)
	at com.test.FruitDAOServiceImpl.fruitCount(FruitDAOServiceImpl.java:682)
	at jrockit.reflect.VirtualNativeMethodInvoker.invoke(Ljava.lang.Object;[Ljava.lang.Object;)Ljava.lang.Object;(Unknown Source)
	at java.lang.reflect.Method.invoke(Ljava.lang.Object;[Ljava.lang.Object;I)Ljava.lang.Object;(Unknown Source)
	at org.springframework.aop.support.AopUtils.invokeJoinpointUsingReflection(AopUtils.java:304)
	at org.springframework.aop.framework.ReflectiveMethodInvocation.invokeJoinpoint(ReflectiveMethodInvocation.java:172)
	at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:139)
	at org.springframework.transaction.interceptor.TransactionInterceptor.invoke(TransactionInterceptor.java:107)</code>

While there could be many reasons because of which this error comes. For me, the problem was that the parameter list I was passing was empty.

I was getting all the fruits for which fruit id is in the fruitIdList I was passing as the query parameter. If the fruitIdList is emply list then this problem comes. I solved it by conditionally framing query based on the list. That is instead of

public Long fruitCount(List<Long> fruitIdList){
	........
	Query fruitQuery = getSession().createQuery("select count(*)  from com.test.Fruits as fruit where fruit.fruitId in (:fruitIdList) order by fruit.fruitName desc ");
	fruitQuery.setParameterList("fruitIdList", fruitIdList);
	List<Long> fruitCountList = fruitQuery.list();
	..........
}

I did this

public Long fruitCount(List<Long> fruitIdList){
	........
	StringBuffer queryString = new StringBuffer("select count(*)  from com.test.Fruits as fruit ");

	if(fruitIdList !=null && !fruitIdList.isEmpty()){
		queryString.append(" where fruit.fruitId in (:fruitIdList) ");
	}

	queryString.append(" order by fruit.fruitName desc ");

	Query fruitQuery = getSession().createQuery(queryString.toString());
	if(fruitIdList !=null && !fruitIdList.isEmpty()){
		fruitQuery.setParameterList("fruitIdList", fruitIdList);
	}

	List<Long> fruitCountList = fruitQuery.list();
	..........
}

In short, append the collection parameter in the query and the query parameter only if the collection is non-empty.

Sorting Hibernate Set using a comparator

Collection mappings in Hibernate can be configured to sort using a specific comparator.

Consider the following mapping

<hibernate-mapping>
<class name="com.example.Person" table="PERSON">
......

<set name="bankAccounts" lazy="true" cascade="all,delete-orphan" inverse="true" sort="natural">
	<key>
		<column name="BANK_ACCOUNT_ID" precision="29" scale="0" not-null="true" />
	</key>
	<one-to-many class="com.example.BankAccount" />
</set>

......
</class>
</hibernate-mapping>

In the example above Hibernate will order the bankAccounts set according to their natural order. If the BankAccout class implements Comparable interface, compareTo method will be used to sort the Set.

What if you want to sort the set differently? What if you want to use a comparator to sort the set.
The sort attribute can have three possible values

The first option is

 <set name="bankAccounts" lazy="true" cascade="all,delete-orphan" inverse="true" sort="natural">

which sorts by natural order

the second option is

 <set name="bankAccounts" lazy="true" cascade="all,delete-orphan" inverse="true" sort="unsorted">

which tells Hibernate that the Set needs not to be sorted

and the third option is

 <set name="bankAccounts" lazy="true" cascade="all,delete-orphan" inverse="true" sort="comparatorClass">

where you specify what comparator should be used to sort the Set.

Let’s see it through code example. For example suppose class BankAccount is defined as below

package com.example;
//import statements

public class BankAccount implements Comparable<BankAccount>, Serializable{

private String accountName;
private String accountShortName;
//getter and setter methods

    public int compareTo(BankAccount that) {

        final int BEFORE = -1;
        final int AFTER = 1;

        if (that == null) {
            return BEFORE;
        }

        Comparable thisAccountName = this.getAccountName();
        Comparable thatAccountName = that.getAccountName();        

        if(thisAccountName == null) {
        	return AFTER;
        } else if(thatAccountName == null) {
        	return BEFORE;
        } else {
        	return thisAccountName.compareTo(thatAccountName);
        }
    }
}

In this case sort=”natural” will sort using the above compareTo method which compares the accountName to sort.
Now suppose you want that for this specific mapping you want the collection to be sorted by, say, accountShortName. You will define a Comparator like this

package com.example;
//import statements
public class BankAccountShortNameComparator implements Comparator<BankAccount>{
	public int compare(BankAccount o1, BankAccount o2) {
		if(o1!=null && o2!=null && o1.getAccountShortName()!=null && o2.getAccountShortName()!=null) {
			return o1.getAccountShortName().compareTo(o2.getAccountShortName());
		}
		if(o1!=null && o2!=null) {
			return o1.compareTo(o2);
		}

		return 0;
	}
}

Then your mapping will say something like this

	<set name="bankAccounts" lazy="true" cascade="all,delete-orphan" inverse="true" sort="com.example.BankAccountShortNameComparator">

In this way Hibernate will use your custom Comparator to sort the set

org.hibernate.TransientObjectException Revisited

This is continuation of my earlier post on org.hibernate.TransientObjectException. Let me write the scenario again

<class name=”com.xxx.A” table=”A” schema=”TESTSCHEMA”>
<id name=”aId” type=”java.lang.Long”>
<column name=”A_ID” precision=”29″ scale=”0″ />
</id>

…………………..
some more mapping elements
…………………..
…………………..
<many-to-one name=”bId” class=”com.xxx.B” fetch=”select”>
<column name=”B_ID” precision=”29″ scale=”0″ />
</many-to-one>

…………………..

A is referring to B using a primary key column of bId of B.

In that post I have mentioned that if B is a transient object and you don’t want to persist the value of B to A then just tell the hibernate to ignore that value by saying
update=”false” insert=”false” in the many to one mapping.

But what if you want to persist the value of foreign key in A.

Then the approach is different. You have to make sure that instance B is persistent not transient.

That is if your code says something like

A a = new A();
B b = new B();
a.setB(b);
…..
…..

session.save(a);

you are in trouble. Because B is in transient state. You have to attach b to the session.
There may be other ways of attaching this transient object to session. The approach I am following is simple. I am reading the value of B from the database using Hibernate. That way, hibernate attaches B to session and it is then a persistent object.

That is I do something like

A a = new A();
B b = session.get(B.class, new Long(1));
a.setB(b);
…..
…..

session.save(a);

I won’t get any exception because this time there is no transient object to save. All the objects are persistent.

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MyEclipse Hibernate Spring tutorial – Managing Hibernate transaction in Spring

Spring transaction management, Hibernate transaction management in Spring

There was a problem/defect in the MyEclipse tutorial on Hibernate Spring. It was not a major problem. The tutorial demonstrates the Spring and Hibernate functionality pretty well. The only problem was that the code does seems to work properly. Because of absence of proper transaction management in code the data was not getting persisted in the database. Because of the caching in Hibernate it seems to the user that the data is being written to the database and then read back.

There are two solutions to the problem. One is to write the transaction management code in the java class itself. Another is to manage transaction via Spring’s transaction management. The second solution makes more sense because it shows of the Spring’s capability of managing transaction in Hibernate. You can see both the solutions here.

I am writing the Spring’s configuration for Hibernate transaction management again.

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE beans PUBLIC "-//SPRING//DTD BEAN//EN" "http://www.springframework.org/dtd/spring-beans.dtd">

<beans>
<bean id="sessionFactory"
class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
<property name="configLocation"
value="file:src/hibernate.cfg.xml">
</property>
</bean>
<bean id="userDAOTarget" class="com.myeclipse.hibernatespring.UserDAO">
<property name="sessionFactory">
<ref bean="sessionFactory" />
</property>
</bean>
<bean id="userDAOService" class="org.springframework.transaction.interceptor.TransactionProxyFactoryBean">
<property name="transactionManager" ref="transactionManager"/>
<property name="transactionAttributes">
<props>
<prop key="add*">PROPAGATION_REQUIRED</prop>
<prop key="update*">PROPAGATION_REQUIRED</prop>
<prop key="delete*">PROPAGATION_REQUIRED</prop>
</props>
</property>
<property name="target"><ref local="persistenceLayer"/></property>
</bean>

<bean id="persistenceLayer"
class="com.myeclipse.hibernatespring.PersistenceLayer"
abstract="false" singleton="true" lazy-init="default"
autowire="default" dependency-check="default">
<property name="userDAO">
<ref bean="userDAOTarget" />
</property>
</bean>
<bean id="transactionManager" class="org.springframework.orm.hibernate3.HibernateTransactionManager">
<property name="sessionFactory"><ref bean="sessionFactory"/></property>
</bean>
</beans>

org.hibernate.TransientObjectException: object references an unsaved transient instance – save the transient instance before flushing:

I was getting this error today.
org.hibernate.TransientObjectException: object references an unsaved transient instance – save the transient instance before flushing:This is the problemTable A was referring to the talbe B using a foreign key

hbm.xml for A looked like this

<class name=”com.xxx.A” table=”A” schema=”TESTSCHEMA”>
<id name=”aId” type=”java.lang.Long”>
<column name=”A_ID” precision=”29″ scale=”0″ />
</id>…………………..
some more mapping elements
…………………..
…………………..
<many-to-one name=”bId” class=”com.xxx.B” fetch=”select”>
<column name=”B_ID” precision=”29″ scale=”0″ />
</many-to-one>

…………………..

I just want to save A with some new value and A will refer to B using some foreign key. But in my transaction I am not going to change B and save the changes. Therefore the many-to-one mapping above is wrong. It does not tell hibernate that class b is immutant. That is it is not going to change. Hibernate thinks that the referred table is not saved therefore it complains and says

org.hibernate.TransientObjectException: object references an unsaved transient instance – save the transient instance before flushing:

Please note that in this case B is transient object. If you don’t really want to save the value of B(transient object) in A , then to get rid of this problem change the many-to-one mapping as

update=”false” insert=”false” fetch=”select”>

But if you want to save the value of B then see my other post

That is, tell hibernate that you are not going to change the values of B, you are just fetching it.

 

ORA-01747: invalid user.table.column, table.column, or column specification

Today I got this error.

ORA-01747: invalid user.table.column, table.column, or column specification.

There could be many reasons why you can get this error. But the basic reason is the query which is finally getting executed to your database is wrong in some way. In my case, I had a column with column name as DESC. I got rid of this problem by simply dropping the table and creating a new one with column name DESCRIPTION instead of DESC. You may not be able to do that if the table is already existing with many records in it.

Other reason could be that, if you are using named query your named query is not syntactically right.

No matter how the you get this error or for that matter any DB error in Hibernate, the best approach is to enable the show_sql variable in hibernate configuration and get the final query which hibernate is firing against the database.

Get this query and try running it directly in your database client. You can easily find out the culprit part of the query from there. Once you find out root cause of the problem you can think of workarounds or solution for this problem.

And yes, don’t forget to turn off the show_sql parameter after you are done with it. It will unnecessarily clutter your logs.

org.hibernate.QueryParameterException: could not locate named parameter

When you query the database in Hibernate using NamedQuery you may run into this error

org.hibernate.QueryParameterException: could not locate named parameter [parm1]

Chances are that you are either trying to set a parmeter to the query which infact doesn’t exist in your query. For example

you are trying to do

query.setParameter(“parm1″, new Long(parm1Value));

whereas, in query there is no parameter like parm1

i.e. query might be just “from Item item”

Or you might be committing some spelling mistake in your code

Eg. query.setParameter(“pram1″, new Long(parm1Value));

and query is “from Item item item.desc like :parm1″

Do note that your parameter name in Java Class is misspelt as pram1 instead of parm1.