How query return multiple values in spring

Question

Please I have this error

"org.hibernate.NonUniqueResultException: query did not return a unique result: 2"

It works perfectly when I have one result and I need to show the two(or more) results found but I don't know how!

Here is my code:

 public class ContactImplDataBase implements ContactDAO {
    //.......
    ...
    public Contact getContactByType(String type) {

    Session session = HibernateUtil.getSessionFactory().openSession();
     Transaction tx = session.beginTransaction();
     Criteria criteria = session.createCriteria(Contact.class)
         .add(Restrictions.like("type", type));
     tx.commit();  
     return (Contact)criteria.uniqueResult();}
    }

And:

 public class ContactImpl implements ContactDAO {
    //...
    ..
    @Override
    public Contact getContactByType(String type) {
        Contact contact=null;
        for(Contact c:contacts){
            if(c.getType().equals(type)){
            contact=c;
            break;
            }
        }
        return contact;
      }
    ...}

And in the controller:

 @RequestMapping(value="/rechercheContact")
    public String rechercheContact(Model model, @RequestParam(value="type")         String type){
        List<Contact> liste=new ArrayList<Contact>();
        liste.add(services.getContactByType(type));

        model.addAttribute("listeContact", liste);
        model.addAttribute("type", type);

        return "ex";
     }

Any help ?!!

Aeseir · Accepted Answer · 2015-08-27 03:22:24Z

1

Your function here:

public Contact getContactByType(String type) {

    Session session = HibernateUtil.getSessionFactory().openSession();
     Transaction tx = session.beginTransaction();
     Criteria criteria = session.createCriteria(Contact.class)
         .add(Restrictions.like("type", type));
     tx.commit();  
     return (Contact)criteria.uniqueResult();} // TAKE NOTE OF THIS
    }

You have set it to only return single unique result.

So that means if there is no single unique result it will thrown an exception.

You have two options here:

Ensure data you searching for is via a unique identifier
Change your getContactByType function to return a List rather than Contact and change (Contact) criteria.unique() to criteria.list();

answered Aug 27, 2015 at 3:22

Aeseir

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Comments

Mathias Begert · Accepted Answer · 2015-08-27 05:08:32Z

Let's do a little code review here:

Format your code for readability.
getContactByType is wrong because there could be more than one contact for one type, so rename your method to getContactsByType and let it return a List<Contact> instead of a single Contact. This has to be changed also in the non provided interface ContactDAO.
The class ContactImpl does not make any sense to me.
Don't use transactions for database reads.

Finally you should get it to work:

public class ContactImplDataBase implements ContactDAO {

    @Override
    public List<Contact> getContactsByType(String type) {
        Session session = HibernateUtil.getSessionFactory().openSession();
        Criteria criteria = session.createCriteria(Contact.class)
            .add(Restrictions.like("type", type));  
        return (List<Contact>)criteria.list();
    }
}

Your controller:

@RequestMapping(value="/searchContacts")
public String searchContacts(Model model, @RequestParam(value="type") String type) {
    List<Contact> list = services.getContactsByType(type);

    model.addAttribute("contactList", list);
    model.addAttribute("type", type);

    return "ex";
}

alex · Accepted Answer · 2015-08-27 03:30:43Z

0

Error message appears to be a hibernate query data return error, presumably the trigger:

return (Contact)criteria.uniqueResult();}

Show that your contact table uses the "type" field query to return multiple records, so there are two ways to solve the solution:

criteria.uniqueResult() => if (Criteria.list() > 0) return criteria.list().get(0)
Modify the number of your database records, maintain

select count(*) contact where type='xxx'

returned 1 records, more than other data deleted. Can also be used to index the database.

answered Aug 27, 2015 at 3:30

alex

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