【发布时间】:2021-05-14 17:54:55
【问题描述】:
我真的是使用 JPA 的新手,但仍然不知道如何实现这一点,我很乐意提供任何建议或解决方案。
Mysql 表:
CREATE TABLE REAL_STATE_TYPES (
real_state_type_id INT(6) NOT NULL,
real_state_type VARCHAR(50) NOT NULL,
PRIMARY KEY (real_state_type_id)
);
CREATE TABLE REAL_STATES (
address VARCHAR(30) NOT NULL,
admin_id VARCHAR(15) NOT NULL,
real_state_type_id INT(6) NOT NULL,
block VARCHAR(3) NOT NULL,
internal_id INT(5) NOT NULL,
PRIMARY KEY (address, block, internal_id),
FOREIGN KEY (real_state_type_id) REFERENCES REAL_STATE_TYPES (real_state_type_id),
FOREIGN KEY (admin_id) REFERENCES ADMINS (admin_id)
);
所以 REAL_STATE 的 id 是 REAL_STATE_TYPE,如果我想保持这个表的关系,我如何通过 JPA 将它们映射到 java 代码中。
我的 java 类是:
import javax.persistence.Enumerated;
import javax.persistence.Id;
import javax.persistence.OneToOne;
import javax.persistence.Table;
@Entity
@Table(name= "real_state_types")
public class RealStateType implements Serializable{
private static final long serialVersionUID = 5442745398170857199L;
@Id
@Column(name= "real_state_type_id ")
private int realStateTypeId;
@Column(name= "real_state_type")
@Enumerated(EnumType.STRING)
private RealStateEnum realStateType;
@OneToOne(mappedBy = "realStateType", cascade = {CascadeType.ALL})
private RealState realState;
还有:
import javax.persistence.Column;
import javax.persistence.EmbeddedId;
import javax.persistence.Entity;
import javax.persistence.Enumerated;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;
@Entity
@Table(name = "real_states")
public class RealState {
@EmbeddedId
private RealStateID realStateID;
@Column(name = "real_state_type_id")
private RealStateType realStateType;
@ManyToOne
@JoinColumn(name = "admin_id")
private Admin admin;
}
【问题讨论】:
标签: java mysql hibernate jpa enums