在jpa标准中,"如果至少有一行返回true"

Question

我正在尝试使用JPA(eclipselink)中的条件api创建后续句子,它很简单地询问某些类别中是否存在某些用户

我想要的句子:

SELECT 
  CASE 
     WHEN EXISTS
          (SELECT * FROM user WHERE category = ?)
     THEN true
     ELSE false
  END 
bind => [10]

我尝试使用此代码:

CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Boolean> criteriaQuery = criteriaBuilder.createQuery(Boolean.class);
Root<T> root = criteriaQuery.from(tclass);

Subquery<T> subquery = criteriaQuery.subquery(tclass);
Root<T> subroot = subquery.from(tclass);
subquery.select(subroot);

Predicate subPredicate = criteriaBuilder.equal(subroot.get("category"),category);
subquery.where(subPredicate);

Predicate predicateExists = criteriaBuilder.exists(subquery);

Case<Boolean> booleancase = criteriaBuilder.<Boolean>selectCase();
Expression<Boolean> booleanExpression =
    booleancase.when(predicateExists,true)
    .otherwise(false);

criteriaQuery.select(booleanExpression);

TypedQuery<Boolean> typedQuery = entityManager.createQuery(criteriaQuery);
typedQuery.getResultList();

可悲的是我的句子是跟随,我想删除最后一个"来自用户":

SELECT 
  CASE 
      WHEN EXISTS
          (SELECT ? FROM user t1 WHERE (t1.category = ?))  
     THEN ?
     ELSE ? 
  END
FROM user t0    
bind => [1, 110, true, false]

任何的想法？

Answer 1

最后我解决了：P

TypedQuery<Boolean> typedQuery = 
     entityManager.createQuery(criteriaQuery).setMaxResults(1);

然后是句子（不一样，但非常接近）：

SELECT 
  CASE 
      WHEN EXISTS
          (SELECT ? FROM user t1 WHERE (t1.category = ?))  
     THEN ?
     ELSE ? 
  END
FROM user t0 LIMIT ?, ? 
bind => [1, 110, true, false,0,1]