序列生成器和USERTYP

Question

我有一个搜索发生器

<id name="idPersistent" type="objectID" column="aa_id">
         <generator class="sequence">
            <param name="sequence">lab_seq</param>
         </generator>
      </id>  

.

objectID是我自己的usertype。 我收到错误

 javax.persistence.PersistenceException: org.hibernate.id.IdentifierGenerationException: this id generator generates long, integer, short or string

.

我该怎么办？

Answer 1

try these links, I hope these will help you

• ID Generator bug in JPA implementation
• Hibernate – Many-To-Many Example – Join Table + Extra Column (Annotation)