我正在尝试动态构造查询,我的下一个目标是添加JOIN子句(我不知道如何使用API).
例如,到目前为止,这段代码对我有用:
...
Class baseClass;
...
CriteriaBuilder cb = JpaHandle.get().getCriteriaBuilder();
CriteriaQuery cq = cb.createQuery(this.baseClass);
Root entity_ = cq.from(this.baseClass);
Predicate restrictions = null;
...
restrictions = cb.conjunction();
restrictions = cb.and(restrictions, entity_.get("id").in(this.listId));
...
cq.where(restrictions);
...
Query qry = JpaHandle.get().createQuery(cq);
(注意:JpaHandle来自wicket-JPA实现)
我的愿望是添加JOIN子句(尽可能通用)!
我在类中有特定的注释(this.baseClass)
例如 :
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "assay_id", nullable = false)
那么,在标准JPA中有没有这样的方法呢？(注意:这不编译)
这是一个实际的失败方法:
...
Join<Experiment,Assay> experimentAssays = entity_.join( entity_.get("assay_id") );
或者像那样:
...
CriteriaQuery<Customer> q = cb.createQuery(Customer.class);
Root<Customer> c = q.from(Customer.class);
SetJoin<Customer, PurchaseOrder> o = c.join(Customer_.orders);
对我来说,如果它可能更加通用,它会很棒......:
...
Join joinClause = …我有一个Service包含tags以下集合的域：
@Entity
public class Service extends AbstractEntity<Long> {
private static final long serialVersionUID = 9116959642944725990L;
@ElementCollection(fetch = FetchType.EAGER, targetClass = java.lang.String.class)
@CollectionTable(name = "service_tags", joinColumns = @JoinColumn(name = "s_id"))
@Column(name = "tag")
private Set<String> tags;
}
我想选择Service小号with特定KEY的Service.tags。
hql接合Service到Service.tags是如下：
select s from Service s INNER JOIN s.tags t where s.status=0 and (s.serviceType=9 or t.tag in ('College'))
但是，上面的 hql 返回给我以下异常：
Caused by: java.lang.IllegalArgumentException: org.hibernate.QueryException: cannot …