@Antoniossss
记录将被插入到Wslog表中,但错误输出为Emp。它还试图插入对象wslog。
-D
WsLog(主)
insert into wslog (totrecords, logid) values (?, ?)
insert into emp (sid, logid, **wslog**, eid, ename) values (?, ?, ?, ?, ?, ?, ?)
EMPS(儿童)
@Column(name = "logid") //has sequence
private long logid;
@Column(name = "totrecords")
private long totrecords;
@OneToMany(mappedBy="wslog")
private List<Emps> emplist;
Repositiory
@Id
@Column(name = "sid") //has sequence
private long sid;
@Column(name = "logid")
private long logid;
@Column(name = "eid")
private Long eid;
@Column(name = "ename")
private String ename;
@ManyToOne(cascade = CascadeType.ALL)
@JoinColumn(name = "logid")
private Wslog wslog;
服务实施
public interface WsLogRepository extends CrudRepository<WsLog, Long> {
}
答案 0 :(得分:0)
您的映射错误,
<强> EMP 强>
@ManyToOne
@JoinColumn(name = "logid")
private WsLog log;
<强> WsLog 强>
@OneToManny(mappedBy="log)
private List<Emp> employees;
这就是你如何建立关系
Employee emp=new Employee();
emp.setLog(log); // log is of type WsLog ofcourse
你没有设置手动ID等。
还要在标识符字段上使用@Id
注释。