JPA 2.0:计算任意的CriteriaQuery?

我写了一个实用类, JDAL J请执行以下操作:

• 计算结果: Long count = JpaUtils.count(em, criteriaQuery);
• 复制标准查询: JpaUtils.copyCriteria(em, criteriaQueryFrom, criteriaQueryTo);
• 获取计数条件: CriteriaQuery<Long> countCriteria = JpaUtils.countCriteria(em, criteria)

等等。。。

如果您对源代码感兴趣,请参见 JpaUtils.java

我已经使用cb.createQuery()对其进行了分类(没有result type参数):

public class Blah() {

    CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
    CriteriaQuery query = criteriaBuilder.createQuery();
    Root<Entity> root;
    Predicate whereClause;
    EntityManager entityManager;
    Class<Entity> domainClass;

    ... Methods to create where clause ...

    public Blah(EntityManager entityManager, Class<Entity> domainClass) {
        this.entityManager = entityManager;
        this.domainClass = domainClass;
        criteriaBuilder = entityManager.getCriteriaBuilder();
        query = criteriaBuilder.createQuery();
        whereClause = criteriaBuilder.equal(criteriaBuilder.literal(1), 1);
        root = query.from(domainClass);
    }

    public CriteriaQuery<Entity> getQuery() {
        query.select(root);
        query.where(whereClause);
        return query;
    }

    public CriteriaQuery<Long> getQueryForCount() {
        query.select(criteriaBuilder.count(root));
        query.where(whereClause);
        return query;
    }

    public List<Entity> list() {
        TypedQuery<Entity> q = this.entityManager.createQuery(this.getQuery());
        return q.getResultList();
    }

    public Long count() {
        TypedQuery<Long> q = this.entityManager.createQuery(this.getQueryForCount());
        return q.getSingleResult();
    }
}

我所做的工作类似于一个criteria builder的构建器,您可以在其中构建一个查询,并使用相同的条件限制调用list()或count()

上面的解决方案都不适用于EclipseLink 2.4.1,它们都以笛卡尔积(N^2)作为结束,这是EclipseLink的一个小技巧,唯一的缺点是我不知道如果你从多个实体中选择,会发生什么,它会试图从你的CriteriaQuery的第一个找到的根中计数,不过,这个解决方案不适用于Hibernate(JDAL可以,但JDAL不适用于EclipseLink)

public static Long count(final EntityManager em, final CriteriaQuery<?> criteria)
  {
    final CriteriaBuilder builder=em.getCriteriaBuilder();
    final CriteriaQuery<Long> countCriteria=builder.createQuery(Long.class);
    countCriteria.select(builder.count(criteria.getRoots().iterator().next()));
    final Predicate
            groupRestriction=criteria.getGroupRestriction(),
            fromRestriction=criteria.getRestriction();
    if(groupRestriction != null){
      countCriteria.having(groupRestriction);
    }
    if(fromRestriction != null){
      countCriteria.where(fromRestriction);
    }
    countCriteria.groupBy(criteria.getGroupList());
    countCriteria.distinct(criteria.isDistinct());
    return em.createQuery(countCriteria).getSingleResult();
  }

你在找这样的东西吗?

/**
 * Counts the number of results of a search.
 * 
 * @param criteria The criteria for the query.
 * @return The number of results of the query.
 */
public <T> Long findCountByCriteria(CriteriaQuery<?> criteria) {
    CriteriaBuilder builder = em.getCriteriaBuilder();

    CriteriaQuery<Long> countCriteria = builder.createQuery(Long.class);
    Root<?> entityRoot = countCriteria.from(criteria.getResultType());
    countCriteria.select(builder.count(entityRoot));
    countCriteria.where(criteria.getRestriction());

    return em.createQuery(countCriteria).getSingleResult();
}

// a search based on the Criteria API
CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<Person> criteria = builder.createQuery(Person.class);
Root<Person> personRoot = criteria.from(Person.class);
criteria.select(personRoot);
Predicate personRestriction = builder.and(
    builder.equal(personRoot.get(Person_.gender), Gender.MALE),
    builder.equal(personRoot.get(Person_.relationshipStatus), RelationshipStatus.SINGLE)
);
criteria.where(personRestriction);
//...

// and to get the result count of the above query
Long count = findCountByCriteria(criteria);

注:我不知道这是否是实现这一点的正确/最佳方法,仍然在学习标准API。。。

条件查询的整体思想是它们是强类型的。因此,在使用原始类型(在CriteriaQuery或根或根中没有泛型)的每个解决方案中,这些解决方案都违背了这个主要思想。我只是遇到了同样的问题,我正在努力用一种“恰当的”(与JPA2一起)方式解决它。

如果您想要所有元素的结果和计数,比如Spring Data的 Page -元素可以执行两个查询。您可以做的是将条件与查询执行分离。

按城市查找用户的示例

 public List<User> getUsers(int userid, String city, other values ...) {

    CriteriaBuilder cb = em.getCriteriaBuilder();
    CriteriaQuery<User> q = cb.createQuery(User.class);
    Root<User> c = q.from(User.class);

    List<Predicate> conditions = createConditions(c, cb, userid, city, ...other values);
    List<User> users = em.createQuery(q.select(c).where(conditions.toArray(new Predicate[] {})).distinct(true))
            .setMaxResults(PAGE_ELEMENTS).setFirstResult(page * PAGE_ELEMENTS).getResultList();
    return users;
}

除了getUser方法之外,您还可以构建第二个方法来计算元素

public Long getElemCount(int userid,  String city, ...other values) {

    CriteriaBuilder cb = em.getCriteriaBuilder();
    CriteriaQuery<Long> q = cb.createQuery(Long.class);
    Root<Location> root = q.from(Location.class);

    List<Predicate> conditions = createConditions(root, cb, userid, page, city, filter, module, isActive);
    Long userCount = em.createQuery(q.select(cb.count(root)).where(conditions.toArray(new Predicate[] {})).distinct(true))
            .getSingleResult();

    return userCount;
}

createConditions方法将同时处理这两个问题,因此不必为条件复制逻辑。

<T> List<Predicate> createConditions(Root<T> root, CriteriaBuilder cb, int userid, String city, ... other values) {

    Join<User, SecondEntity> usr = root.join("someField");
    // add joins as you wish

    /*
     * Build Conditions
     */
    List<Predicate> conditions = new ArrayList<>();

    conditions.add(cb.equal(root.get("id"), userid));

    if (!city.equals("")) {
       conditions.add(cb.like(...));
    }

   // some more conditions...

    return conditions;
}

long elementCount=yourCriteriaClassInstance.getElementCount(…); List users=yourCriteriaClassInstance.getUsers(…)

我用冬眠和 criteria api

public Long getRowsCount(List<Criterion> restrictions ) {
       Criteria criteria = getSession().createCriteria(ThePersistenclass.class);
       for (Criterion x : restrictions)
             criteria.add(x);
   return criteria.setProjection(Projections.rowCount()).uniqueResult();        

}

希望帮助