html - 加入游戏框架

Question

我没用过jpa，想问一下有没有join

如果我这样做

String queryStringGet="SELECT c.* FROM Category c "+

"LEFT JOIN User us ON us.id = c.id "+

"+"WHERE c.user_id="+id;

List<Object[]> list = JPA.em().createNativeQuery(queryString).getResultList();

有效但作为对象出现，我不知道如何在 View 中列出 :(

由于 play.db.jpa.Model 不支持我该如何加入？以及如何列出对象？

谢谢

Answer 1

JPQL 不支持联接。有一个 explanation of how to do it here .

您可以像这样在列表中获取结果:

String hql = "SELECT c FROM Category c "
           + "LEFT JOIN User us WITH us.id = c.id "
           + "WHERE c.user_id= :userId";
Query query = JPA.em().createQuery(hql, Categories.class);
query.setParameter("userId", userId);
List<Category> categories = query.getResultList();