org.hibernate.MappingException：找不到具有逻辑名称的列

Posted 2023-02-25

【中文标题】org.hibernate.MappingException：找不到具有逻辑名称的列

【英文标题】：org.hibernate.MappingException: Unable to find column with logical name

【发布时间】：2012-01-27 07:40:36

【问题描述】：

您好，我的表格如下：

1- medical_company：

account_entity

表

account_entity_id

列上的medical_company_id

外键

（不是pk） 列 1 列2 第三栏

2- account_entity：

account_entity_id

(pk) 列 1 列2 第三栏

3- 人：

person_id

(pk) 列 1 列2 第三栏

4- employee_company：

medical_company_id

上的

medical_company

表上的 company_id 外键

person_id

上的

person

表上的 employee_id 外键 列 1 列2

实体：

1- 医疗公司：

@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- 人：

 @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- 员工公司：

@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;
    

在尝试运行应用程序时，我收到以下错误：

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)

请告知我为什么会收到此错误，以及如何解决它。

【问题讨论】：

解决了吗？我也有同样的情况。

【参考方案1】：

此错误告诉您，Medical_company 表中没有名为 medical_company_id 的列。 medical_company 上的列仅称为 id。

【讨论】：

但是这个名字有一个外键！ @JoinColumn(name = "medical_company_id"

【参考方案2】：

我将删除 MedicalCompany 中的 referencedColumnName 属性，因为您正在命名 AccountEntity 的主键字段。我认为只有当它引用非主键字段时才有必要。

@JoinColumn(name = "medical_company_id", referencedColumnName = "account_entity_id")

【讨论】：

这是我遇到的问题（在 Hibernate 6.0.13.Final 上）

太棒了！也为 Hibernate 5.3.1 解决了我的问题。但是，当我开始在我的 persistence.xml 中使用 hibernate.globally_quoted_identifiers 属性时，我才开始看到这个问题。不知道为什么。

也解决了我的问题！！这应该被标记为答案！

【参考方案3】：

对于遇到此问题并使用globally_quoted_identifiers 的人，这对我有帮助：

改变

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

到

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

注意包装referencedColumnName 值的反引号。

【参考方案4】：

我在不同的情况下收到相同的错误。我尝试用字符串键加入 2 个表。此代码块发送相同的错误。

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

我通过添加列和 Fetch 类型解决了问题（我在实体中有 2 个连接，所以我需要添加 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;