sin*_*pop 6 java orm hibernate jpa eclipselink
我想坚持一个拥有一些资源(内联或附件)的邮件实体.首先,我将它们视为双向关系:
@Entity
public class Mail extends BaseEntity {
    @OneToMany(mappedBy = "mail", cascade = CascadeType.ALL, orphanRemoval = true)
    private List<MailResource> resource;
    private String receiver;
    private String subject;
    private String body;
    @Temporal(TemporalType.TIMESTAMP)
    private Date queued;
    @Temporal(TemporalType.TIMESTAMP)
    private Date sent;
    public Mail(String receiver, String subject, String body) {
        this.receiver = receiver;
        this.subject = subject;
        this.body = body;
        this.queued = new Date();
        this.resource = new ArrayList<>();
    }
    public void addResource(String name, MailResourceType type, byte[] content) {
        resource.add(new MailResource(this, name, type, content));
    }
}
@Entity
public class MailResource extends BaseEntity {
    @ManyToOne(optional = false)
    private Mail mail;
    private String name;
    private MailResourceType type;
    private byte[] content;
}
当我救了他们:
Mail mail = new Mail("asdasd@asd.com", "Hi!", "...");
mail.addResource("image", MailResourceType.INLINE, someBytes);
mail.addResource("documentation.pdf", MailResourceType.ATTACHMENT, someOtherBytes);
mailRepository.save(mail);
执行了三个插入:
INSERT INTO MAIL (ID, BODY, QUEUED, RECEIVER, SENT, SUBJECT) VALUES (?, ?, ?, ?, ?, ?)
INSERT INTO MAILRESOURCE (ID, CONTENT, NAME, TYPE, MAIL_ID) VALUES (?, ?, ?, ?, ?)
INSERT INTO MAILRESOURCE (ID, CONTENT, NAME, TYPE, MAIL_ID) VALUES (?, ?, ?, ?, ?)
然后我认为仅使用OneToMany关系会更好.无需在每个MailResource中保存哪个Mail:
@Entity
public class Mail extends BaseEntity {
    @OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
    @JoinColumn(name = "mail_id")
    private List<MailResource> resource;
    ...
    public void addResource(String name, MailResourceType type, byte[] content) {
        resource.add(new MailResource(name, type, content));
    }
}
@Entity
public class MailResource extends BaseEntity {
    private String name;
    private MailResourceType type;
    private byte[] content;
}
生成的表完全相同(MailResource有一个FK到Mail).问题是执行的SQL:
INSERT INTO MAIL (ID, BODY, QUEUED, RECEIVER, SENT, SUBJECT) VALUES (?, ?, ?, ?, ?, ?)
INSERT INTO MAILRESOURCE (ID, CONTENT, NAME, TYPE) VALUES (?, ?, ?, ?)
INSERT INTO MAILRESOURCE (ID, CONTENT, NAME, TYPE) VALUES (?, ?, ?, ?)
UPDATE MAILRESOURCE SET mail_id = ? WHERE (ID = ?)
UPDATE MAILRESOURCE SET mail_id = ? WHERE (ID = ?)
为什么这两个更新?我正在使用EclipseLink,使用另一个JPA提供程序作为Hibernate,这种行为是否相同?哪种解决方案更好?
更新: - 如果我不使用@JoinColumn,EclipseLink会创建三个表:MAIL,MAILRESOURCE和MAIL_MAILRESOURCE.我认为这是完全逻辑的.但是对于@JoinColumn,它只有足够的信息来创建两个表,在我看来,只进行插入,没有更新.
映射定义了关系的拥有方,因此对于 JPA,它提供了更好的方法来处理关联。Join Column 仅定义关系列。由于 JPA 是完全基于反射的框架,我可以想到对 Mapped by 进行的优化,因为通过这种方式很容易找到拥有方。
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