JPA 不会通过包装对象将 Id 添加到数据库

Question

使用 JPA 我有一个数据配置，它可以毫无问题地添加到数据库中：

   @Entity @Table(name = "dataConfigurations", uniqueConstraints = {@UniqueConstraint(columnNames = {"name"})})
    @NamedQueries({@NamedQuery(name = "DataConfigurations.names", query = "SELECT d.name FROM DataConfiguration d"), @NamedQuery(name = "DataConfiguration.findByName", query = "SELECT d FROM DataConfiguration d WHERE d.name = :name")})
    public class DataConfiguration implements Serializable
    {

        @Transient
        private static final long serialVersionUID = 1L;
        @Id //@GeneratedValue(strategy = GenerationType.)
            //@TableGenerator(name = "dataConfigurationSequence", table = "Sequences", pkColumnName = "NameSequences", valueColumnName = "valueSequence", pkColumnValue = "dataConfigurationId")
        private long id;
        @OneToMany(mappedBy = "dataConfiguration", cascade = CascadeType.ALL)
        private List<Content> contents; // "Data" is a reserved SQL word.
        @ManyToMany(mappedBy = "dataConfigurations", cascade = CascadeType.ALL)
        private List<LabelType> labelTypes;
        private String name;
        @OneToMany(mappedBy = "dataConfiguration", cascade = CascadeType.ALL
        private List<Template> templates;

此数据配置包含一个内容列表（一对多关系）， 其他映射（模板和 labelTypes 现在并不重要，但稍后在我的程序中，它们可能会给我带来与内容相同的问题）

@Entity @Table(name = "contents", uniqueConstraints = {@UniqueConstraint(columnNames = {"dataConfigurationId", "displayOrder"})})
    public class Content implements Serializable
    {
        @Transient
        private static transient final long serialVersionUID = 1L;
        @Id //@GeneratedValue(strategy = GenerationType.IDENTITY)
            //@TableGenerator(name = "contentSequence", table = "Sequences", pkColumnName = "NameSequences", valueColumnName = "valueSequence", pkColumnValue = "contentId")
        private long id;
        private String content; // "Data" is a reserved word in traditional SQL.
            @ManyToOne(optional = false, fetch = FetchType.LAZY, cascade = CascadeType.ALL) @JoinColumn(name = "dataConfigurationId")
        private DataConfiguration dataConfiguration;
        private int displayOrder;
        private int editableFrom;
        private String typeDiscriminator;

这是启动保存方法的控制器方法：

public String saveDataConfiguration()
{
    dataConfiguration = new DataConfiguration(getNextId(), dataConfigurationName, contents);
    dataConfigurationFacade.create(dataConfiguration);
    return null;
}

创建方法是一部分AbstractFacade 的

public void create(T entity)
{
    getEntityManager().persist(entity);
        getEntityManager().flush();
}

当我执行该方法时，除了内容表中数据配置的 Id 之外的所有内容都会被保存。 （数据配置的外键不保存在内容中​​） 任何提示和/或技巧都会非常有帮助！

Answer 1

您可能忘记初始化每个内容的 dataConfiguration 字段。这是映射到连接列的列，因此 Hibernate 将检查它以确定两个实体之间是否存在关联。