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如何将JPQL查询分配给一个简单的Java对象?

jpql java-ee jpa sql

1

ErVeY · 技术社区 · 15 年前

我需要将jpql结果的结果分配给一个简单的类java对象

class myObject() {
@id
private Long id;
private String Name;
private String description;
...
//getters and setters
}

我需要以某种方式存储这个SQL查询的结果,例如

//可能是任何一张桌子从anyTable中选择DISTINCT c.table\u id、c.name、NULL作为description

如何在JPQL中执行此操作,然后将结果分配给我的对象?

2 回复 | 直到 15 年前

1

Pascal Thivent 15 年前

这个问题非常不清楚。所以这里有一个模糊的答案:

EntityManagerFactory emf = Persistence.createEntityManagerFactory("MyPu");
EntityManager em = emf.createEntityManager();    
Query q = em.createQuery("select f.id, f.name from Foo f");
List<Object[]> foos = (List<Object([]>)q.getResultList();

MyObject o = new MyObject();
o.setFoos(foos);

2

0

Slava 13 年前

我想,这就是答案!

SELECT NEW entities.PersonWell(c.peopleId, c.name, c.age, c.height, c.weight)
FROM 
  People AS c;

public PersonWell(Integer peopleId, String name, short age, short height, short weight, short speechspeed) {
        this.peopleId = peopleId;
        this.name = name;
        this.age = age;
        this.height = height;
        this.weight = weight;
        this.speechspeed = speechspeed;
    }

List<PersonWell> resultList = query.getResultList();

希望对你有帮助!:)