Question

我在尝试坚持一对多关系时遇到异常。我可以看到问题发生的原因是在插入子（链接）时没有使用父级（策略）自动生成的id。

SQL命令：

CREATE TABLE POLICIES
(
  ID INTEGER NOT NULL PRIMARY KEY GENERATED ALWAYS AS IDENTITY (START WITH 1, INCREMENT BY 1),
  CURRENTLINKID INTEGER,
);

CREATE TABLE LINKS
(
  ID INTEGER NOT NULL PRIMARY KEY GENERATED ALWAYS AS IDENTITY (START WITH 1, INCREMENT BY 1),
  ORDINAL INTEGER NOT NULL,
  LINK VARCHAR(50) NOT NULL,
  POLICYID INTEGER NOT NULL
);

ALTER TABLE POLICIES
ADD FOREIGN KEY(CURRENTLINKID) 
REFERENCES LINKS(ID);

ALTER TABLE LINKS
ADD FOREIGN KEY(POLICYID) 
REFERENCES POLICIES(ID);

实体：

@Entity
@Table(name = "POLICIES", catalog = "", schema = "")
public class PolicyEntity
{
    ...
    @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, mappedBy = "_policy")
    private Collection<LinkEntity> _linkEntityCollection;
    ...
}

@Entity
@Table(name = "LINKS", catalog = "", schema = "")
public class LinkEntity
{
    ...
    @JoinColumn(name = "POLICYID", referencedColumnName = "ID")
    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL, optional = false)
    private PolicyEntity _policy;
    ...
}

创建实体并坚持：

LinkEntity link = new LinkEntity();
link.setLink("link");
link.setOrdinal(1);

List<LinkEntity> links = new ArrayList<LinkEntity>();
links.add(link);

PolicyEntity policy = new PolicyEntity();
policy.setLinks(links);

EntityManager entityManager = _emf.createEntityManager();

// Begin transaction
entityManager.getTransaction().begin();

// persist
entityManager.persist(policy);

// Commit the transaction
entityManager.getTransaction().commit();

// Close this EntityManager
entityManager.close();

例外：

javax.persistence.RollbackException：Exception [TOPLINK-4002]（Oracle TopLink Essentials - 2006.8（Build 060830））： oracle.toplink.essentials.exceptions.DatabaseException内部例外：java.sql.SQLIntegrityConstraintViolationException：Column ＆＃39;策略ID＆＃39;不能接受NULL值。错误代码：20000调用：INSERT INTO LINKS（LINK，ORDINAL，POLICYID）VALUES（？，？，？）bind =＆gt; [链接， 1， null ] 查询：InsertObjectQuery（xxx.xxxx.xxx.xxx.datalayer.entities.LinkEntity@1bb813b）

致电：插入链接（链接，序列，政策）价值（？，？，？） bind =＆gt; [link，1，null]＆lt; ---如何在这里设置自动生成的策略ID？

Answer 1

根据克里斯的评论更新。

我意识到我需要在链接实体中实际设置策略。现在链接插入毫无例外。

PolicyEntity policy = new PolicyEntity();
LinkEntity link = new LinkEntity();
link.setLink("link");
link.setOrdinal(1);
link.setPolicy(policy) // set policy so it wasn't null

List<LinkEntity> links = new ArrayList<LinkEntity>();
links.add(link);

policy.setLinks(links);

JPA toplink cascade持久化，javadb具有null外键

1 个答案: