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

Question

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.

it was solved? I have same situation.

verystrongjoe
– verystrongjoe

2015-09-16 10:01:40 +00:00
Commented Sep 16, 2015 at 10:01 — verystrongjoe
– verystrongjoe, Commented Sep 16, 2015 at 10:01

Mark Rotteveel · Accepted Answer · 2014-09-25 11:22:08Z

44

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")

edited Sep 25, 2014 at 11:22

Mark Rotteveel

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answered Dec 25, 2011 at 17:48

BikeHikeJuno

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3 Comments

Romeo Sierra Over a year ago

This is the problem that I was having (on Hibernate 6.0.13.Final)

Naxi Over a year ago

Awesome ! Solved my issue as well for Hibernate 5.3.1. However, I only started seeing this issue as soon as I started using hibernate.globally_quoted_identifiers property in my persistence.xml. Not sure why.

Stacie Over a year ago

Solved my issue as well!! This should be marked as the answer!

Alex Barnes · Accepted Answer · 2011-12-24 15:45:28Z

12

This error is telling you that there is no column on the medical_company table called medical_company_id. The column on medical_company is just called id.

answered Dec 24, 2011 at 15:45

Alex Barnes

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1 Comment

Mahmoud Saleh Over a year ago

but there's a foreign key with this name ! @JoinColumn(name = "medical_company_id"

Talha Dilber · Accepted Answer · 2021-06-03 08:08:28Z

1

I received same error for different situation. I try to join 2 table with String key. This code block send same error.

@OneToMany
@JoinColumn(name = "SHIPMENT_ID", referencedColumnName = "PRODUCT_SHIPMENT_GROUP")
private List<ProductShipment> productShipments = new ArrayList<>();

I solved question with adding column and Fetch type (I have 2 join in the entity so I need to add FetchMode)=>

@OneToMany
@JoinColumn(name = "SHIPMENT_ID", referencedColumnName = "PRODUCT_SHIPMENT_GROUP")
@Fetch(value = FetchMode.SUBSELECT)
private List<ProductShipment> productShipments = new ArrayList<>();

@Column(name = "PRODUCT_SHIPMENT_GROUP")
private String productShipmentGroup;

answered Jun 3, 2021 at 8:08

Talha Dilber

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Comments

Jdslk · Accepted Answer · 2022-11-23 09:32:23Z

0

Make sure your parent entity's primary key name: @Column(name = "id") is the same name as the referencedColumnName (foreign key) in the child entity: @JoinColumn(name = "created_by", referencedColumnName = "id")

I.e: id = id

answered Nov 23, 2022 at 9:32

Jdslk

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Comments

Sattam AlShoubaki · Accepted Answer · 2023-05-22 12:05:41Z

0

This might happen if you removed column from entity but column referenced by another entity

answered May 22, 2023 at 12:05

Sattam AlShoubaki

11 bronze badge

Comments

Jan Mares · Accepted Answer · 2020-11-27 22:30:22Z

-1

For those who ran into this problem and are using globally_quoted_identifiers, this helped me:

Change

@JoinColumns({
    @JoinColumn(name = "new_col_1", referencedColumnName = "ref_col_1"),
    @JoinColumn(name = "new_col_2", referencedColumnName = "ref_col_2")
})

To

@JoinColumns({
    @JoinColumn(name = "new_col_1", referencedColumnName = "`ref_col_1`"),
    @JoinColumn(name = "new_col_2", referencedColumnName = "`ref_col_2`")
})

Notice the backtick wrapping the referencedColumnName value.

answered Nov 27, 2020 at 22:30

Jan Mares

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