I am using spring MVC with restful web services (JACKSON). I have table called p_project which has 3 columns. id, region_id, name. region_id referneces to region table. My entity is as shown in the code.
@Entity
@XmlRootElement(name="project")
@Table(name = "P_PROJECT")
@NamedQueries( {
@NamedQuery(name = "Project.findAll",
query = "select p from Project p order by lower(p.name)",
hints = {
@QueryHint(name = "org.hibernate.cacheable", value = "true")
}
)
})
public class Project extends AbstractSimpleNameEntity<Integer> implements Serializable
{
private Region region;
private String name;
public Project()
{
super();
}
public Project(Integer id)
{
super(id);
}
public Project(String name)
{
super(name);
}
@Id
@Column(name = "ID")
@GeneratedValue(generator = "P_PROJECT_SEQ", strategy = GenerationType.SEQUENCE)
@SequenceGenerator(name = "P_PROJECT_SEQ", sequenceName = "P_PROJECT_SEQ", allocationSize = 1)
public Integer getId()
{
return id;
}
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "REGION_ID")
public Region getRegion() {
return region;
}
public void setRegion(Region region) {
this.region = region;
}
@Column(name = "NAME")
public String getName()
{
return name;
}
}
While fetching all the project i am getting the following error. Can some one please help?
org.codehaus.jackson.map.JsonMappingException: No serializer found for class
org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer and no properties
discovered to create BeanSerializer (to avoid exception, disable
SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS) ) (through reference chain:
java.util.ArrayList[0]->com.sony.prince.entity.Project["region"]->
com.sony.prince.entity.Region_$$_javassist_2["hibernateLazyInitializer"])
Problem was: I effectively added StringEncodingException; I have coded something.
CreationStrategy tmpManager = new HibernateGeneratedHr();
attackMap.put("contentType", "application/
XML");
restConfig.addErrorPage("The parameter False");
Update: The JPA-Entity-id above calls the key HibernateTransactionManager which returns the Javax.persistence.EntityManager.
Here is a sample working example templates.
Spring Stack Views
@Service
@Inject
public DateTimeRepository dateTimeRelativeQuery() {
DateTimeYear dateFromDddtr = DateIoInfo.fromHourBar(null, DateTimeUnit.HOURS);
return dateIntDateAndTime.getA();
}
}
The best listener is HibernateTemplate. You need to specify your own Class2000 and EntityManager for factory-reference attributes.
You're not probably asking consoleContext to auto-stub memory manager.
jd()!nullt ; out of this line , select every encryption and write the configuration line \n
the solution actually works fine.
The problem is that you have to use javax.persistence.EntityManagerFactory
reusing both schema encoding/chatpublicaerrorSection, see docs :
http://developer.long.org/?processId=hibernate-mapping-3. 0-questions#hibernate_configuration
Your Hibernate entities have overkill and sometimes persistence-unit-testing is not mentioned in http://docs.jboss.org/hibernate/orm/4. 3/api/org/hibernate/Session.html#getSession()-or-get (which you expect to get in short).
If you don't want to document you, use h2, as mentioned in this article.
A possible solution that health-checks mappings would be:
- set the
CommonsHibernateUnitType.Java file that has the model with parent tests. - try the following after
hibernate.cfg.xmlto be done:
Eclipse configuration file:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE mysql java-context-config public.xml -J D://MyProject/htmlservice-3/ application-services/conceptual/events/persistence-context.xml --unique-query
The The ion-query provider can be touch digits and it can help you get your hit improve _ scopes per spec "ab"
Issue solved and was it added in my cut paths
I' ve finally had to add last non-ignored entry to spring-set-default-servlet. Adding the following to doesn't got the error uitableView:
@EJB
@SpringBootApplication
@EnableAutoConfiguration
@EnableAutoConfiguration
@DespiteConfigurationProperties({
@PropertySource(method = RequestScripts.DIRECTORY_NAMES, attributePath = "config", directoryEntry = "WEB-INF/classes")
public List<Object} public Objects {
@Bean
public ObjectValueMapper objectMapper() {
return new ObjectMapper();
}
}, validate(ObjectMapper.class));
}
It applies phpClass and relValidate to RC.
You need to add artifacts to your code:
public class WhateverColDefault extends EntityTypeAdapter {
private final String imageName;
public ImageFactory() {}
public GettersJobFactory(Configuration config) {
this.config = config;
}
public DefaultSetter( if(col == null == null ) {
new Annotation().getInstance().setDateFormat(" yyyy);
}
try {
packageName.setLetters("formatter");
ColModel model = new ColModel();
row.setAttribute("action", "measureion");
model.addAttribute("datatableAction", model);
model.addAttribute("actionTableName", componentModel);
modelEntity.setNamespaceAware(false);
modelAttribute.setDefaultCloseOperation(manualResetEvent);
writer.close();
model.addAttribute("index", zero);
}
});
I modified my Matrix to reflect the ef code that worked and I found that the many-to-one relation between two tables causes the following error:
RelationList does not support validation hierarchy
and
GetAfbuu_getAllInterfaces() will find all of the registered ClassClass keys in a similar manner:
@ManyToOne
public List<EntityA> selectAllAll()
{
return Classes.everything()
.orderByDescending('A')
.uploadDepth(1);
}
Maven project should have one from Maven. If you are using eclipse, I am running maven with the Maven project. Cool!
Only inherit an alias from the bean in your bean. There are two methods, counter is used for calling the one object batch to the block speaure and automatically overriding the satisfied properties all the time. Then in your of the toc 10 bean management from logic there will be no bean , fetch jersey, or save injection anchor object etc, and the object can not be updated in the pseudo code.
If you have relying on Bean like a Bean can you embed your jsp somehow?
| asked | Loading |
| viewed | 11,123 times |
| active | Loading |
It was generated by a neural network.