JPA EntityManager createQuery() error

Question

This is failing:

public List<TypeActionCommerciale> requestTypeActionCommercialeSansNip() throws PersistenceException {
    Query query = createQuery("from TypeActionCommercialeImpl where type != :type1");
    query.setParameter("type1", TypeActionCommercialeEnum.NIP);
    return (List<TypeActionCommerciale>) query.list();
}

exception:

Hibernate: select typeaction0_.id as id1_102_, typeaction0_.libelle as libelle3_102_, typeaction0_.code as code4_102_, typeaction0_.type as type5_102_ from apex.typeActionCommerciale typeaction0_ where typeaction0_.type<>?

ERROR org.hibernate.engine.jdbc.spi.SqlExceptionHelper.logExceptions(SqlExceptionHelper.java:129)

No value specified for parameter 1 org.hibernate.exception.DataException: could not extract ResultSet at

i use setProperties but i have the same error:

public List<TypeActionCommerciale> requestTypeActionCommercialeSansNip() throws PersistenceException {
    Query query = createQuery("from TypeActionCommercialeImpl where type  <> :type1");
    final Map<String, Object> properties = new HashMap<>();
    properties.put("type1", TypeActionCommercialeEnum.NIP);
    query.setProperties(properties);
    return (List<TypeActionCommerciale>) query.list();
}

The query is good. The problem states that parameter 1 is missing. Where is TypeActionCommercialeEnum.NIP? It is like it is not in the classpath. Can you try "from TypeActionCommercialeImpl where type != TypeActionCommercialeEnum.NIP" (and remove the setParameter line) to see what error you get — Ian Mc
– Ian Mc, Commented Jan 25, 2016 at 21:58
@IanMc error: org.hibernate.QueryException: Unable to resolve path [TypeActionCommercialeEnum.NIP], unexpected token [TypeActionCommercialeEnum] [from com.metier.impl.TypeActionCommercialeImpl where type != TypeActionCommercialeEnum.NIP] — Mercer
– Mercer, Commented Jan 25, 2016 at 22:02

Andrea Catania · Accepted Answer · 2016-01-25 23:04:43Z

1

The problem is here query.setParameter("type1", TypeActionCommercialeEnum.NIP);

The enum type is not defined in hibernate so you must store the name of enum and use it for the query (the easy way) then use:

query.setString("type1", TypeActionCommercialeEnum.NIP.name());

To use enum directly (the hard way) you must implement your CustomUserType . You can find here how https://docs.jboss.org/hibernate/orm/5.0/manual/en-US/html/ch06.html#types-custom

The main advantages of use CustomUserType are:

you can store into DB an integer (that is more smaller) instead of a string that represents the enum.
Delegate the parsing to hibernate during storing and retrieving of object.
You can use the enum directly into the query (like you are trying to do)

edited Jan 25, 2016 at 23:04

answered Jan 25, 2016 at 21:46

Andrea Catania

1,3833 gold badges21 silver badges38 bronze badges

Sign up to request clarification or add additional context in comments.

10 Comments

Andrea Catania Over a year ago

What type is the field "type"?

Mercer Over a year ago

type character varying(255)

Andrea Catania Over a year ago

The problem is that the keyword "type" is private keyword. Can you try to change the field name for a test please?

Mercer Over a year ago

it's work under Hibernate 3 here i'm under Hibernate 5

Andrea Catania Over a year ago

it's very strange problem... try to use setString() and paste the stack trace

|

Abdelhak · Accepted Answer · 2016-01-25 22:17:23Z

0

Try to use <> instead of != like this:

  "from TypeActionCommercialeImpl where type <> :type1"

edited Jan 25, 2016 at 22:17

answered Jan 25, 2016 at 21:34

Abdelhak

8,4154 gold badges25 silver badges37 bronze badges

2 Comments

Luan Oliveira Over a year ago

Use setProperties method Instead setParameter. And change != to <>

Mercer Over a year ago

The method getResultList() is undefined for the type Query

Mercer · Accepted Answer · 2016-01-27 22:22:37Z

i resolve my pb i have a class public class EnumUserType<E extends Enum<E>> implements UserType and i implement this method:

@Override
public Object nullSafeGet(ResultSet rs, String[] names, SessionImplementor session, Object owner)
        throws HibernateException, SQLException {
     String name = rs.getString(names[0]);  
        Object result = null;  
        if (!rs.wasNull()) {  
            result = Enum.valueOf(clazz, name);  
        }  
        return result;
}

@Override
public void nullSafeSet(PreparedStatement st, Object value, int index, SessionImplementor session)
        throws HibernateException, SQLException {
    if (null == value) {  
        st.setNull(index, Types.VARCHAR);  
    } else {  
        st.setString(index, ((Enum<E>) value).name());  
    }  

}

Collectives™ on Stack Overflow

JPA EntityManager createQuery() error

3 Answers 3

10 Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

10 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related