session filter can't work:query must begin with SELECT or FR

Yashnoo · **Joined:** Wed Dec 17, 2003 4:24 am **Posts:** 188

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2.1.6:

<class name="com.test.bean.Broker" table="broker" dynamic-update="true" dynamic-insert="true">
<id name="broker_id" column="broker_id" type="string">
<generator class="com.test.util.HibernateGenerator"/>
</id>
<property name="city">
<column name="city" sql-type="varchar(20)"/>
</property>

<set name="brokers" inverse="true" lazy="true">
<key column="broker_id"/>
<one-to-many class="com.fangtoo.bean.Broker"/>
</set>
</class>:

Session s=HibernateUtil.currentSession();
Transaction t=s.beginTransaction();

Broker boss=(Broker)s.get(Broker.class,"743");
Set brokers=boss.getBrokers();
log.info("boss is: "+brokers);
Broker bb=(Broker)s.filter(brokers,"this.broker_type='boss'");
log.info("boss name is : "+bb.getBrief_name());
t.commit();
HibernateUtil.closeSession();

12:38:33,953 INFO TestHiber:50 - boss is: [com.fangtoo.bean.Broker@691dee]
net.sf.hibernate.QueryException: query must begin with SELECT or FROM: this.broker_type [this.broker_type='boss']
at net.sf.hibernate.hql.ClauseParser.token(ClauseParser.java:84)
at net.sf.hibernate.hql.PreprocessingParser.token(PreprocessingParser.java:123)
at net.sf.hibernate.hql.ParserHelper.parse(ParserHelper.java:29)
at net.sf.hibernate.hql.QueryTranslator.compile(QueryTranslator.java:149)
at net.sf.hibernate.hql.QueryTranslator.compile(QueryTranslator.java:138)
at net.sf.hibernate.hql.FilterTranslator.compile(FilterTranslator.java:26)
at net.sf.hibernate.impl.SessionFactoryImpl.getFilter(SessionFactoryImpl.java:306)
at net.sf.hibernate.impl.SessionImpl.getFilterTranslator(SessionImpl.java:3456)
at net.sf.hibernate.impl.SessionImpl.filter(SessionImpl.java:3513)
at net.sf.hibernate.impl.SessionImpl.filter(SessionImpl.java:3408)
at com.fangtoo.tests.TestHiber.main(TestHiber.java:51)
Exception in thread "main"

MySQL 4.01 innoDB:

no

INFO:

Hi everyone:

I want to filter a collection that querey from database. But the filter report error:

Code:

query must begin with SELECT or FROM: this.broker_type [this.broker_type='boss']

I have seen Hibernate In Action. It write filter code in this way:

Code:

Object o=session.filter(parent.getChildSet(),"this.user_name='lyo'");

But why it can't work in my code ? Thks!

Yashnoo · **Joined:** Wed Dec 17, 2003 4:24 am **Posts:** 188

I find that if the code is:

Code:

Object bb=s.createFilter(brokers,"this.broker_type='boss'");

It can work well and not exception.
But if the code is the following , it will report Exception:

Code:

Object bb=s.createFilter(brokers,"this.broker_type='boss'").uniqueRequest();

Does "uniqueResult" can't be used with hibernate filter?

Yashnoo · **Joined:** Wed Dec 17, 2003 4:24 am **Posts:** 188

any idea?

mgreer · **Posted:** Thu Feb 24, 2005 9:48 pm

Filters filter collections, as I understand it. Why would you use a filter to get a unique result? If you know the criteria to get a single row, use that instead of a filter.

Yashnoo · **Joined:** Wed Dec 17, 2003 4:24 am **Posts:** 188

mgreer wrote:

Filters filter collections, as I understand it. Why would you use a filter to get a unique result? If you know the criteria to get a single row, use that instead of a filter.

Thks for your reply. But I want to know if hibernate filter can do this. And why it throws Exception. Sometimes it is important to filter a unique result from a collection.