Question

我与以下实体共同创建了“世界展望组织”项目:事实和项目。

项目内容提及事实。

@RooJavaBean
@RooToString
public class ItemFacture {

    @ManyToOne
    private Facture facture;
...

这是《事实法典》:

@RooJavaBean
@RooToString
@RooEntity
public class Facture {

    private String nom;
    private String type;
}

直到此为止。我想创建一种习俗,选择包含具体事实的所有项目:

@SuppressWarnings("unchecked")
public static List<ItemFacture> findByFacture(Facture facture) {

    Query q = entityManager().createQuery("SELECT o FROM ItemFacture AS o WHERE o.facture = :facture");
    q.setParameter("facture", facture);

    return q.getResultList();
}

当我试图处决这一寻找者时,这给我留下了这一错误:

服务器(Error):现场“装置”在测试中不存在。项目设计或没有持久性; nes例外是 j, 持久性:在 com测试中不存在实地“装置”。项目内容

I created a few custom finders and every one of them worked just fine except this one.

I tried adding @Persistent to the field and @PersistenceCapable(identityType = IdentityType.APPLICATION) to Facture but it still doesn t work.

Anyone know what is the problem?

Answer 1

In AppEngine, references are made with the ID of the referenced object. So event if ItemFacture contain and attribute facture, it s actually stored as factureId.

因此,书写方法是:

@SuppressWarnings("unchecked")
public static List<ItemFacture> findByFacture(Facture facture) {

    Query q = entityManager().createQuery("SELECT o FROM ItemFacture AS o WHERE o.factureId = :facture");
    q.setParameter("facture", facture.getId());

    return q.getResultList();
}

Answer 2

i 猜测为什么不做这样的工作,即项目类型不是“Entity”,因此日本邮局没有这方面的想法。加上“RooEntity roo”的“Facture”类别,产生了一个项目“Factures_Roo_Entity.aj(STS在成套探索器中以违约方式隐藏这些档案),把“Entity annotation on You 项目Facture-class与使该项目成为积极记录的其他部分相联(http://martinfowler.com/eaaCatalog/activeRecord.html)

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