我认为这是您想要使用的条件组合（顺便说一句，帮助定义Hibernate实体bean比帮助定义表结构更容易）：

String[] employeeNames = { "test", "test2" };
List<Survey> surveys = getSession().createCriteria(Survey.class).add(
        Restrictions.and
        (
            Restrictions.eq("surveyNumber", 1),
            Restrictions.in("employeeName", employeeNames)
        )
    ).list();

List<EmployeeSurvey> employeeSurvey = getSession().createCriteria
(EmployeeSurvey.class).add(
    Restrictions.and
    (
        Restrictions.eq("survey_no", 1), 
        Restrictions.in("employee_name", new String[] {"test", "test1"})
    )).list();

public List<Breed> getManyBreeds(StringBuilder breedIds)  {
        System.out.println(breedIds);
        SessionFactory factory = HibernateConnection.getConnection();
        Session session = factory.openSession();
        Criteria criteria = session.createCriteria(Breed.class);
        criteria.add(Restrictions.in("id", breedIds.toString()));
        List<Breed> breeds = criteria.list();
        session.close();
        return breeds;
    }

我遇到了同样的问题，我已将所有ID存储在stringBuilder中，但当我尝试此方法时，它会给我一个错误，没有返回任何值，请帮助我解决此问题。