JPA 2.0:计数任意 CriteriaQuery? [英] JPA 2.0: count for arbitrary CriteriaQuery?

问题描述

我正在尝试实现以下便捷方法:

I am trying to implement the following convenience method:

/**
 * Counts the number of results of a search.
 * @param criteria The criteria for the query.
 * @return The number of results of the query.
 */
public int findCountByCriteria(CriteriaQuery<?> criteria);

在 Hibernate 中，这是由

In Hibernate, this is done by

criteria.setProjection(Projections.rowCount());

JPA 中与上面的等价物是什么?我找到了许多简单的计数示例，但没有一个使用 CriteriaQuery 应确定其行数.

What is the equivalent to the above in JPA? I found numerous simple count examples, but none of them made use of a CriteriaQuery whose row count should be determined.

不幸的是，我发现@Pascal 的答案不正确.这个问题很微妙，只有在你使用连接时才会出现:

I unfortunately found out that @Pascal's answer is not the correct one. The problem is very subtle and only shows up when you use joins:

// Same query, but readable:
// SELECT *
// FROM Brain b
// WHERE b.iq = 170

CriteriaQuery<Person> query = cb.createQuery(Person.class);
Root<Person> root = query.from(Person.class);
Join<Object, Object> brainJoin = root.join("brain");
Predicate iqPredicate = cb.equal(brainJoin.<Integer>get("iq"), 170);
query.select(root).where(iqPredicate);

当调用 findCountByCriteria(query) 时，它会因以下异常而死亡:

When calling findCountByCriteria(query), it dies with the following exception:

org.hibernate.hql.ast.QuerySyntaxException: Invalid path: 'generatedAlias1.iq' [select count(generatedAlias0) from xxx.tests.person.dom.Person as generatedAlias0 where generatedAlias1.iq=170]

还有其他方法可以提供这样的 CountByCriteria 方法吗?

Is there any other way to provide such a CountByCriteria method?

推荐答案

我写了一个实用类，JDAL JpaUtils去做:

I wrote a utility class, JDAL JpaUtils to do it:

• 计数结果:Long count = JpaUtils.count(em,criteriaQuery);
• copy CriteriaQueries:JpaUtils.copyCriteria(em,criteriaQueryFrom,criteriaQueryTo);
• 获取计数条件:CriteriaQuerycountCriteria = JpaUtils.countCriteria(em,criteria)

等等...

如果你对源代码感兴趣，请看JpaUtils.java

If you are interested in the source code, see JpaUtils.java

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