使用JPA Criteria-API从连接表中选择对象

Question

我正在与以下问题争吵两天,希望你能给我一个正确的方向.我在研究过程中发现的教程和示例总是只显示如何在标准api中轻松加入工作.首先,我有两个班:

@Entity
public class Offer {

  private String name;
  @ManyToOne private Location location;
  private String tags;
}

和

@Entity
public class Location {
  private String name;
  private string tags;
}

因为我需要避免循环引用,所以这些类之间的连接只是单向的.这个类中有很多其他属性,我想根据我的搜索过滤器构建动态查询.以下SQL语句将解释我喜欢做什么:

SELECT l 
FROM Offer o 
JOIN o.location l
WHERE o.tags LIKE :sometag AND l.tags LIKE :someothertag

用标准api实现这个后,我得到了这段代码:

CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
CriteriaQuery<Location> criteriaQuery = criteriaBuilder.createQuery(Location.class);
criteriaQuery = criteriaQuery.distinct(true);
Join location;
ArrayList<Predicate> whereList = new ArrayList<Predicate>();

// if filter by offer, use offer as main table and join location table
if (filter.getOfferTags() != null) {
  Root<Offer> offer = criteriaQuery.from(Offer.class);

  location = offer.join("location");


  // limit to offering tags
  Path<String> tagPath = offer.get("tags");

  for (String tag : filter.getOfferTags()) {
    Predicate whereTag = criteriaBuilder.like(tagPath, "%" + tag + "%");
    whereList.add(whereTag);
  }

} else {
  // else use location table as base
  location = (Join<Location, Location>) criteriaQuery.from(Location.class);
}

但是如果我执行此操作,我会从H2数据库中收到以下错误消息:

Column "LOCATION.ID" not found; SQL statement: 
SELECT DISTINCT LOCATION.ID, LOCATION.NAME
FROM OFFER t0, LOCATION t1 
WHERE t0.TAGS LIKE ? AND t1.TAGS LIKE ?

数据库期望t1.ID和t1.NAME在select子句中而不是LOCATION.ID和LOCATION.NAME.如何告诉JPA创建"正确"的请求？我在代码中遗漏了什么吗？

我正在使用带有Eclipse Link和H2数据库的Glassfish 3.1.1.

Answer 1

我想你错过了查询中的一个选择:

criteriaQuery.select(location);