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Unable to fetch results using createQuery  RSS feed

 
sreenivas jeenor
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Posts: 125
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Hi All.

In my Code i am tyring tring fetch values based on condition.

Here the code
-----------------
import org.hibernate.*;
import org.hibernate.Session;
import org.hibernate.cfg.*;



import java.util.*;

public class SelectHQLExample {

public static void main(String[] args) {
Session session = null;

try{
// This step will read hibernate.cfg.xml and prepare hibernate for use
SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory();
session =sessionFactory.openSession();

String searchString="Car";
String sql="From Insurance p where p.lngInsuranceId :id";
String queryString ="from Insurance p where p.insuranceName like ?";
Query q = session.createQuery(queryString).setString(0,searchString);

for(Iterator it=q.iterate();it.hasNext() {
Object[] row = (Object[]) it.next();
System.out.println("ID: " + row[0]);

}



session.close();
}catch(Exception e){
System.out.println("vasu........."+e.getMessage());
}finally{
}

}
}

-------------------------------
The output what i am getting is
----------------------------

Hibernate: select insurance0_.ID as col_0_0_ from insurance insurance0_ where in
surance0_.insurance_name like ?

Thanks
 
Mark Spritzler
ranger
Sheriff
Posts: 17309
11
IntelliJ IDE Mac Spring
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You are missing the call that runs the query.

q.list() this returns a List of results, but you just create the Query object and try to call q.iterate(), is there even such a method? Anyway, instead of calling q.iterate call

List l = q.list(); and look a the values in l.

Also remove that String sql, because it is not valid HQL in the Where portion of the query.

Mark
 
It is sorta covered in the JavaRanch Style Guide.
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