我通常不熟悉Java,Hibernate或EntityManager。我已经使用了很多NHibernate但是使用Linq和Fluent NHibernate几乎没有必要使用标准api所以我有点挣扎试图让以下查询工作
获取ReferenceCodeMetaData的所有实例,其中siteId == 1和 codeType =='sampleUnits'
我知道这应该是
的内容em.createQuery(Restrictions.and(
Restrictions.eq("siteId", siteId),
Restrictions.eq("codeType", codeType)
))
但我无法正确输入通用字样。有人可以帮我填写这个功能的主体吗?
public ReferenceCodeMetaData[] getMatching(EntityManager em, Integer siteId, String codeType) {
return ...
}
答案 0 :(得分:2)
您正在使用EntityManager中的方法,该方法将CriteriaQuery或JPQL查询作为参数,但您的参数类型的类型是指Hibernate Criteria。以下是此类方法的三种可能实现方式:
public ReferenceCodeMetaData[] getMatchingWithJPQL(EntityManager em,
Integer siteId,
String codeType) {
String jpql = "SELECT r FROM ReferenceCodeMetaData r where siteId = :siteId AND codeType = :codeType";
TypedQuery<ReferenceCodeMetaData> query = em.createQuery(jpql, ReferenceCodeMetaData.class);
query.setParameter("siteId", siteId);
query.setParameter("codeType", codeType);
List<ReferenceCodeMetaData> result = query.getResultList();
return result.toArray(new ReferenceCodeMetaData[result.size()]);
}
public ReferenceCodeMetaData[] getMatchingWithCriteriaAPI(EntityManager em,
Integer siteId,
String codeType) {
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<ReferenceCodeMetaData> cq = cb.createQuery(ReferenceCodeMetaData.class);
Root<ReferenceCodeMetaData> root = cq.from(ReferenceCodeMetaData.class);
cq.select(root)
.where(cb.and(
cb.equal(root.get("siteId"), cb.parameter(Integer.class, "siteId")),
cb.equal(root.get("codeType"), cb.parameter(String.class, "codeType"))));
TypedQuery<ReferenceCodeMetaData> query = em.createQuery(cq);
query.setParameter("siteId", siteId);
query.setParameter("codeType", codeType);
List<ReferenceCodeMetaData> result = query.getResultList();
return result.toArray(new ReferenceCodeMetaData[result.size()]);
}
public ReferenceCodeMetaData[] getMatchingWithJHibernateCriteria(EntityManager em,
Integer siteId,
String codeType) {
HibernateEntityManager hem = em.unwrap(HibernateEntityManager.class);
Session session = hem.getSession();
// If you use some older version of Hibernate, then unwrap method is not
// available and you can use following instead of two lines above:
// Session session = (Session) em.getDelegate();
List<ReferenceCodeMetaData> result = session.createCriteria(ReferenceCodeMetaData.class)
.add(Restrictions.eq("siteId", siteId) )
.add(Restrictions.eq("codeType", codeType))
.list();
return result.toArray(new ReferenceCodeMetaData[result.size()]);
}
前两个是标准JPA,最后一个是Hibernate特定的。所有这些都生成等效的SQL查询并产生相同的结果。在这种情况下,我会选择第一个。