JPA / Hibernate：CriteriaBuilder - 如何使用关系对象创建查询？ [英] JPA / Hibernate: CriteriaBuilder - How to create query using relationship object?

问题描述

我有以下四个表：

SCHEDULE_REQUEST表：
ID，
APPLICATION_ID（FK）

应用表：
ID。

USER_APPLICATION TABLE： APPLICATION_ID（FK），
USER_ID（FK）

用户表：
ID，
NAME

想要创建Criteria构建器，其中条件是为指定的用户ID选择ScheduleRequests。

我有以下代码：

  List< User> usersList = getSelectedUsers（）; // userList包含我想要选择的用户
 
 CriteriaBuilder builder = getJpaTemplate（）。getEntityManagerFactory（）。getCriteriaBuilder（）; 
 CriteriaQuery< ScheduleRequest> criteria = builder.createQuery（ScheduleRequest.class）; 
 Root< ScheduleRequest> scheduleRequest = criteria.from（ScheduleRequest.class）; 
 criteria = criteria.select（scheduleRequest）; 
 
 ParameterExpression< User> 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< ScheduleRequest> query = getJpaTemplate（）。getEntityManagerFactory（）。createEntityManager（）。createQuery（criteria）; 
 
 //编译时间错误在这里：
 // TypedQuery< ScheduleRequest>类型中的方法setParameter（Parameter< T> ;, T） （ParameterExpression< User> ;, List< User>）
 query.setParameter（usersIdsParam，usersList）; 
 
返回query.getResultList（）; 
  

你能帮我一下如何将查询过滤器传递给关系对象吗？
我认为我在application.userApplications.user中所做的是错误的？
请真的需要帮助。
预先感谢您。

解决方案

使用规范Metamodel和一些连接，它应该可以工作。试试如果你从下面的伪代码中得到一些提示（未经测试）：

$ $ $ $ $ $ $ $ $ $ $ = cb.disjunction（）;
if（usersList！= null）{
ListJoin< ScheduleRequest，Application>应用程序= scheduleRequest.join（ScheduleRequest_.applications）;
ListJoin< Application，UserApplication> userApplications = applications.join（Application_.userApplications）;
加入< UserApplication，User> user = userApplications.join（UserApplication_.userId）; （字符串用户名：usersList）{
谓词= builder.or（谓词，builder.equal（user.get（User_.name），用户名））;
。
}
}

criteria.where（predicate）;
...

为了理解Criteria Queries，请看这些教程：
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元模型类，这应该由编译器/ IDE自动构建。

I have the following four tables:

SCHEDULE_REQUEST TABLE: ID, APPLICATION_ID (FK)

APPLICATION TABLE: ID. CODE

USER_APPLICATION TABLE: APPLICATION_ID (FK), USER_ID (FK)

USER TABLE: ID, NAME

Now I wanted to create Criteria builder where condition is to select ScheduleRequests for specified user Ids.

I have the following codes:

List<User> usersList = getSelectedUsers(); // userList contains users I wanted to select

CriteriaBuilder builder = getJpaTemplate().getEntityManagerFactory().getCriteriaBuilder();
CriteriaQuery<ScheduleRequest> criteria = builder.createQuery(ScheduleRequest.class);
Root<ScheduleRequest> scheduleRequest = criteria.from(ScheduleRequest.class);
criteria = criteria.select(scheduleRequest);

ParameterExpression<User> 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<ScheduleRequest> query = getJpaTemplate().getEntityManagerFactory().createEntityManager().createQuery(criteria);

// Compile Time Error here:
// The method setParameter(Parameter<T>, T) in the type TypedQuery<ScheduleRequest> is not // applicable for the arguments (ParameterExpression<User>, List<User>)
query.setParameter(usersIdsParam, usersList);

return query.getResultList();

Can you please help me how to pass query filter to a relationship object? I think what I did in "application.userApplications.user" is wrong? Please really need help. Thank you in advance.

解决方案

Using the canonical Metamodel and a couple of joins, it should work. Try if you get some hints from the following pseudo-code (not tested):

...
Predicate predicate = cb.disjunction();
if (usersList != null) {
    ListJoin<ScheduleRequest, Application> applications = scheduleRequest.join(ScheduleRequest_.applications);
    ListJoin<Application, UserApplication> userApplications = applications.join(Application_.userApplications);
    Join<UserApplication, User> user = userApplications.join(UserApplication_.userId);
    for (String userName : usersList) {
        predicate = builder.or(predicate, builder.equal(user.get(User_.name), userName));
    }
}

criteria.where(predicate); 
...

In order to understand Criteria Queries, have a look at these tutorials: http://www.ibm.com/developerworks/java/library/j-typesafejpa/ http://docs.oracle.com/javaee/6/tutorial/doc/gjitv.html

The second link should also guide you on how to use Metamodel classes, that should be built automatically by the compiler / IDE.

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