本文介绍了org.hibernate.MappingException:无法找到具有逻辑名称的列的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我的表格如下:
1- 医疗公司:
- account_entity 表 account_entity_id 列(不是pk)上的
- medical_company_id 外键
- column1
- column2
- column3
- 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
- account_entity_id (pk)
- column1
- column2
- column3
3- 个人:
- person_id (pk)
- column1
- column2
- column3
- person_id (pk)
- column1
- column2
- column3
4- employee_company:
- medical_company_id 上 medical_company 表上的
- company_id外键 person 上 person 表上的
- employee_id外键
- column1
- column2
- company_id foreign key on medical_company table on medical_company_id
- employee_id foreign key on person table on person_id
- column1
- column2
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- 人员:
@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;
}
}
当尝试运行该应用程序时,出现以下错误:
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)
请告知为什么我会收到此错误,以及如何解决它.
please advise why i am getting this error, and how to solve it.
推荐答案
此错误告诉您在medical_company表上没有名为medical_company_id的列. medical_company上的列仅称为id.
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.
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