JPA / Hibernate:CriteriaBuilder – 如何使用关系对象创建查询?
我有以下四个表:
SCHEDULE_REQUEST表: ID,APPLICATION_ID(FK)
应用程序表: ID,代码
USER_APPLICATION TABLE: APPLICATION_ID(FK),USER_ID(FK)
USER TABLE: ID,NAME
现在我想创建一个CriteriaBuilder
,其条件是为指定的用户ID选择ScheduleRequests
。
我有以下代码:
List usersList = getSelectedUsers(); // userList contains users I wanted to select CriteriaBuilder builder = getJpaTemplate().getEntityManagerFactory().getCriteriaBuilder(); CriteriaQuery criteria = builder.createQuery(ScheduleRequest.class); Root scheduleRequest = criteria.from(ScheduleRequest.class); criteria = criteria.select(scheduleRequest); ParameterExpression usersIdsParam = null; if (usersList != null) { usersIdsParam = builder.parameter(User.class); params.add(builder.equal(scheduleRequest.get("application.userApplications.user"), usersIdsParam)); } criteria = criteria.where(params.toArray(new Predicate[0])); TypedQuery query = getJpaTemplate().getEntityManagerFactory().createEntityManager().createQuery(criteria); // Compile Time Error here: // The method setParameter(Parameter, T) in the type TypedQuery is not // applicable for the arguments (ParameterExpression, List) query.setParameter(usersIdsParam, usersList); return query.getResultList();
你能帮我解决一下如何将查询filter传递给关系对象吗? 我认为我在“application.userApplications.user”中所做的是错误的? 请真的需要帮助。
先感谢您!
使用规范的Metamodel和几个连接,它应该工作。 如果您从以下伪代码(未测试)获得一些提示,请尝试:
... Predicate predicate = cb.disjunction(); if (usersList != null) { ListJoin applications = scheduleRequest.join(ScheduleRequest_.applications); ListJoin userApplications = applications.join(Application_.userApplications); Join user = userApplications.join(UserApplication_.userId); for (String userName : usersList) { predicate = builder.or(predicate, builder.equal(user.get(User_.name), userName)); } } criteria.where(predicate); ...
要了解Criteria Queries,请查看以下教程: http : //www.ibm.com/developerworks/java/library/j-typesafejpa/ http://docs.oracle.com/javaee/6/tutorial/ DOC / gjitv.html
第二个链接还应指导您如何使用Metamodel类,这些类应由编译器/ IDE自动构建。