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Jpa 如何在多租户中设置@SequenceGenerator架构_Jpa_Ejb_Eclipselink_Multi Tenant - Fatal编程技术网

Jpa 如何在多租户中设置@SequenceGenerator架构

Jpa 如何在多租户中设置@SequenceGenerator架构,jpa,ejb,eclipselink,multi-tenant,Jpa,Ejb,Eclipselink,Multi Tenant,我面临这个问题,我的实体档案是: @Entity @Table(name="profile") @Multitenant(value=MultitenantType.TABLE_PER_TENANT) @TenantTableDiscriminator(type=TenantTableDiscriminatorType.SCHEMA, contextProperty="eclipselink.tenant-id") @Inheritance(strategy = InheritanceType.

我面临这个问题,我的实体档案是:

@Entity
@Table(name="profile")
@Multitenant(value=MultitenantType.TABLE_PER_TENANT)
@TenantTableDiscriminator(type=TenantTableDiscriminatorType.SCHEMA, contextProperty="eclipselink.tenant-id")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "type", discriminatorType = DiscriminatorType.STRING, length = 17)
public class Profile implements Serializable, Comparable<Profile> {

    private static final long serialVersionUID = 1L;

    @Id
    @SequenceGenerator(name = "profile_id_generator", sequenceName = "profile_identification_seq", allocationSize = 1, schema = ?)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "profile_id_generator")
    @Column(unique = true, nullable = false)
    @Expose
    private Integer identification;
@实体
@表(name=“profile”)
@多租户(值=多租户类型。每个租户的表)
@TenantTableDiscriminator(type=TenantTableDiscriminatorType.SCHEMA,contextProperty=“eclipselink.tenant id”)
@继承(策略=InheritanceType.SINGLE_表)
@鉴别器列(name=“type”,鉴别器类型=鉴别器类型.STRING,长度=17)
公共类概要文件实现了可序列化、可比较的{
私有静态最终长serialVersionUID=1L;
@身份证
@SequenceGenerator(name=“profile\u id\u generator”,sequenceName=“profile\u identification\u seq”,allocationSize=1,schema=?)
@GeneratedValue(策略=GenerationType.SEQUENCE,generator=“profile\u id\u generator”)
@列(unique=true,nullable=false)
@暴露
私有整数识别;

请问,如何根据当前用户租户id设置@SequenceGenerator注释中的架构?同时,我发现的简单方法是在每个事务开始时更新架构以适合当前租户
将搜索路径设置为租户