CODEWITHSUNDEEP
springhibernate
Lena
Lena

Reputation: 3

How query return multiple values in spring

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 ?!!

Upvotes: 0

Views: 2449

Answers (3)

Mathias Begert
Mathias Begert

Reputation: 2471

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";
}

Upvotes: 0

alex
alex

Reputation: 1

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:

  1. criteria.uniqueResult() => if (Criteria.list() > 0) return criteria.list().get(0)

  2. 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.

Upvotes: 0

Aeseir
Aeseir

Reputation: 8414

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:

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

Upvotes: 1

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