“基本”属性类型不应为持久性实体

java spring-data-jpa spring-boot

2022-09-03 14:21:07

我在另一个实体类中引用了一个实体，并收到此错误。下面是示例代码。我在持久性中也有这些类.xml。

导致此问题的原因是什么？我正在使用Spring数据JPA和Hibernate。

import javax.persistence.*;
@Entity
@Table(name = "users", schema = "university")
public class UsersEntity {
    private long id;

    @JoinColumn(name = "address_id", nullable = false)
    private Address address;

    @Id
    @Column(name = "id")
    public long getId() {
        return id;
    }

    public void setId(long id) {
        this.id = id;
    }

    public Address getAddress() {
        return address;
    }

    public void setAddress(Address address) {
        this.address = address;
    }
}

import javax.persistence.*;
@Entity
@Table(name = "address", schema = "university")
public class AddressEntity {
    private long id;
    private String street;

    @Id
    @Column(name = "id")
    public long getId() {
        return id;
    }

    public void setId(long id) {
        this.id = id;
    }

    @Basic
    @Column(name = "street")
    public String getStreet() {
        return street;
    }

    public void setStreet(String street) {
        this.street = street;
    }
}

答案 1

因此，请尝试以下操作：

@Entity
@Table(name = "users", schema = "university")
public class UsersEntity {
    private Long id;

    private AddressEntity address;

    @Id
    @Column(name = "id")
    public Long getId() {
        return id;
    }

    public void setId(long id) {
        this.id = id;
    }

   @OneToOne 
   @JoinColumn(name = "address_id", nullable = false)
   public AddressEntity getAddress() {
        return address;
    }

    public void setAddress(AddressEntity address) {
        this.address = address;
    }
}


@Entity
@Table(name = "address", schema = "university")
public class AddressEntity {
    private Long id;
    private String street;

    @Id
    @Column(name = "id")
    public Long getId() {
        return id;
    }

    public void setId(long id) {
        this.id = id;
    }


    @Column(name = "street")
    public String getStreet() {
        return street;
    }

    public void setStreet(String street) {
        this.street = street;
    }
}

基本上，我所做的是替换Long的long。并添加了@OneToOne我删除了@Basic，因为它是可选的。我相信它应该只是工作

答案 2

您错过了实体之间的关系类型。例如，您可以将单向@OneToOne定义为

//...
public class UsersEntity {
//..
  @OneToOne(mappedBy = "ugdArea", fetch = FetchType.LAZY)
  @JoinColumn(name = "address_id", nullable = false)
  private Address address;
//...
}