java - JPA-2.0 简单的 Select-Where 问题

Question

我遇到了一个关于 JPA-2.0 关系查询的问题。如何选择任何具有至少一个 type = B 的 Event 的 Dataset？

@Entity
class Dataset {
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "dataset")
    public List<Event> events;
}

@Entity
class Event {
    @ManyToOne
    @JoinColumn
    public Dataset dataset;

    public Type type;
}

enum Type {
     A, B, C
}

我的出发点是

CriteriaBuilder _builder = em.getCriteriaBuilder();
CriteriaQuery<Dataset> _criteria = _builder.createQuery(Dataset.class);

// select from
Root<Dataset> _root = _criteria.from(Dataset.class);
_criteria.select(_root);

// apply some filter as where-clause (visitor)
getFilter().apply(
   _root, _criteria, _builder, em.getMetamodel()
);

// how to add a clause as defined before?
...

对此有任何想法。我尝试创建一个子查询和一个连接，但不知何故我做错了，结果总是得到所有数据集。

Answer 1

尝试

SELECT d FROM DataSet d WHERE EXISTS 
    (SELECT e FROM Event e WHERE e.dataSet = d and e.type = :type)

编辑:正如 Pascal 指出的那样，您似乎正在使用 Criteria API。对此不太熟悉，但我会尝试一下。

CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<Dataset> criteria = builder.createQuery(Dataset.class);

Root<Dataset> root = criteria.from(Dataset.class);
criteria.select(root);

// build the subquery
SubQuery<Event> subQuery = criteria.subQuery(Event.class);
Root<Event> eventRoot = subQuery.from(Event.class);
subQuery.select(eventRoot);

ParameterExpression<String> typeParameter = builder.parameter(String.class);
Predicate typePredicate = builder.equal(eventRoot.get(Event_.type), typeParameter));

// i have not tried this before but I assume this will correlate the subquery with the parent root entity
Predicate correlatePredicate = builder.equal(eventRoot.get(Event_.dataSet), root);
subQuery.where(builder.and(typePredicate, correlatePredicate);

criteria.where(builder.exists(subQuery)));

List<DataSet> dataSets = em.createQuery(criteria).getResultList();

呸，辛苦了。我现在要回到 Linq。