Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

org.hibernate.MappingException: Unable to find column with logical name

Tags:

hi my tables are as follows:

1- medical_company:

  • medical_company_id foreign key on account_entity table account_entity_id column (not a pk)
  • column1
  • column2
  • column3

2- account_entity:

  • account_entity_id (pk)
  • column1
  • column2
  • column3

3- person:

  • person_id (pk)
  • column1
  • column2
  • column3

4- employee_company:

  • company_id foreign key on medical_company table on medical_company_id
  • employee_id foreign key on person table on person_id
  • column1
  • column2

ENTITIES:

1- MedicalCompany:

@SuppressWarnings("serial") @Entity @Table(name = "medical_company") public class MedicalCompany implements Serializable {  @Id @GeneratedValue(strategy = GenerationType.IDENTITY) @Column(name = "id", unique = true, nullable = false) @Basic(fetch = FetchType.EAGER) private Long id;  @OneToOne @Cascade(value = { CascadeType.ALL }) @JoinColumn(name = "medical_company_id", referencedColumnName = "account_entity_id") private AccountEntity accountEntity; 

}

2- AccountEntity:

@SuppressWarnings("serial") @Entity @Table(name = "account_entity") public class AccountEntity implements Serializable {      @Id     @GeneratedValue(strategy = GenerationType.IDENTITY)     @Column(name = "account_entity_id", unique = true, nullable = false)     @Basic(fetch = FetchType.EAGER)     private Long id;  } 

3- Person:

 @SuppressWarnings("serial") @Entity @Table(name = "person") public class Person implements Serializable {      @Id     @GeneratedValue(strategy = GenerationType.IDENTITY)     @Column(name = "person_id", unique = true, nullable = false)     @Basic(fetch = FetchType.EAGER)     private Long id;  } 

4- EmployeeCompanyId:

@SuppressWarnings("serial") @Embeddable public class EmployeeCompanyId implements Serializable {      @ManyToOne     private Person person;      @ManyToOne     private MedicalCompany medicalCompany;      @Size(max = 150, message = "{long.value}")     @Column(name = "title_text", length = 150, nullable = true)     private String titleText;      @Column(name = "employee_manager")     private long employeeManager;   } 

5- EmployeeCompany:

@SuppressWarnings("serial") @Entity @Table(name = "employee_company") @AssociationOverrides(value = {         @AssociationOverride(name = "pk.medicalCompany", joinColumns = @JoinColumn(referencedColumnName = "medical_company_id")),         @AssociationOverride(name = "pk.person", joinColumns = @JoinColumn(referencedColumnName = "person_id")),         @AssociationOverride(name = "pk.titleText"),         @AssociationOverride(name = "pk.employeeManager") }) public class EmployeeCompany implements Serializable {      @EmbeddedId     private EmployeeCompanyId pk = new EmployeeCompanyId();      @Transient     public void setEmployeeManager(long employeeManager) {         this.pk.setEmployeeManager(employeeManager);     }      public long getEmployeeManager() {         return pk.getEmployeeManager();     }      @Transient     public void setTitleText(String titleText) {         this.pk.setTitleText(titleText);     }      public String getTitleText() {         return pk.getTitleText();     }      public void setPerson(Person person) {         this.pk.setPerson(person);     }      @Transient     public Person getPerson() {         return this.pk.getPerson();     }      public void setMedicalCompany(MedicalCompany medicalCompany) {         this.pk.setMedicalCompany(medicalCompany);     }      @Transient     public MedicalCompany getMedicalCompany() {         return this.pk.getMedicalCompany();     }      public void setPk(EmployeeCompanyId pk) {         this.pk = pk;     }      public EmployeeCompanyId getPk() {         return pk;     }  } 

when trying to run the application, i am getting the following error:

org.hibernate.MappingException: Unable to find column with logical name: medical_company_id in org.hibernate.mapping.Table(medical_company) and its related supertables and secondary tables     at org.hibernate.cfg.Ejb3JoinColumn.checkReferencedColumnsType(Ejb3JoinColumn.java:550)     at org.hibernate.cfg.BinderHelper.createSyntheticPropertyReference(BinderHelper.java:126)     at org.hibernate.cfg.ToOneFkSecondPass.doSecondPass(ToOneFkSecondPass.java:110)     at org.hibernate.cfg.AnnotationConfiguration.processFkSecondPassInOrder(AnnotationConfiguration.java:520)     at org.hibernate.cfg.AnnotationConfiguration.secondPassCompile(AnnotationConfiguration.java:380)     at org.hibernate.cfg.Configuration.buildMappings(Configuration.java:1206)     at org.springframework.orm.hibernate3.LocalSessionFactoryBean.buildSessionFactory(LocalSessionFactoryBean.java:717)     at org.springframework.orm.hibernate3.AbstractSessionFactoryBean.afterPropertiesSet(AbstractSessionFactoryBean.java:211)     at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1477)     at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1417) 

please advise why i am getting this error, and how to solve it.

like image 611
Mahmoud Saleh Avatar asked Dec 24 '11 12:12

Mahmoud Saleh


1 Answers

I would remove the referencedColumnName attribute in MedicalCompany because you are naming the primary key field of AccountEntity. I think it is only necessary if you if it references a non-primary-key field.

@JoinColumn(name = "medical_company_id", referencedColumnName = "account_entity_id") 
like image 179
BikeHikeJuno Avatar answered Oct 06 '22 23:10

BikeHikeJuno