Question

我正在尝试使用EclipseLink JPA＆amp; amp;来模拟一个简单的OneToMany关系。 MySQL的。它失败了一些错误消息。任何人都可以帮助我理解它缺少什么吗？

@Entity
public class Job implements Serializable {
    @Id
    @GeneratedValue
    private long jobId;

    @OneToMany(targetEntity=Applicant.class,mappedBy="job",cascade=CascadeType.ALL)
    private List<Applicant> applicant;

    //Gettters and Setters
}

@Entity
public class Applicant {
    @Id
    @GeneratedValue
    private long applicantId;


    @ManyToOne
    @JoinColumn(name="job_Id")
    private Job job;


    //Gettters and Setters
}

TestCode

Job job=new Job ();
List <Applicant> applicants=new ArrayList<Applicant>();
Applicant applicant=new Applicant();
applicants.add(applicant);
job.setApplicant(applicants);
em.persist(job);// EntityManager Reff

错误讯息：

Internal Exception: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'job_Id' in 'field list'

Answer 1

您需要在Applicant表中添加外键列。

您Job与@JoinColumn(name="job_Id")的注释关系。 @JoinColumn不是必需的，因为JPA完全取决于约定优于配置，这意味着如果您不覆盖它们，则会隐含默认值。如果没有此注释，EclipseLink将搜索名为job_jobId（<field name>_<id column name in target table>）的列。由于您声明加入列为job_Id，因此您需要在Applicant表格中包含该列。

这些数据库表应与您的实体映射一起使用（@JoinColumn(name = "job_Id")）：

TABLE JOB
    JOBID (PRIMARY KEY)

TABLE APPLICANT
    APPLICANTID (PRIMARY KEY)
    JOB_ID (FOREIGN KEY TO JOB#JOBID)

Answer 2

@Entity
public class Job implements Serializable{

        @Id
        @GeneratedValue
        @Column(name="job_id")
        private long jobId;

        @OneToMany(mappedBy="job",cascade=CascadeType.ALL)
        private List<Applicant> applicants;

        public Job(){
            applicants = new ArrayList<Applicant>();
        }

        public List<Applicant> getApplicants(){
            //force clients through addApplicant() method to encapsulate setting both sides of relationship
            return Collections.unmodifiableList(applicants);
        }

        public void addApplicant(Applicant applicant){
            //set both sides of relationship
            applicant.setJob(this);
            applicants.add(applicant);
        }
    }

    @Entity
    public class Applicant {
        @Id
        @GeneratedValue
        private long applicantId;


        @ManyToOne
        @JoinColumn(name="job_Id")
        private Job job;
    }   


    Job job = new Job ();
    Applicant applicant = new Applicant();
    job.addApplicant(applicant);
    em.persist(job);

Eclipselink JPA onetomany关系

2 个答案: