JPA NamedQuery 获取不同列

发布于 2024-12-10 12:13:59 字数 581 浏览 0 评论 0原文

我想从我的用户表中获取不同城市的列表。我认为下面的代码可以工作，但给出了一个错误：

User.java
@Entity
@Table(name="user_info")
...
@NamedQuery(name = "User.all.cities", query = "SELECT distinct u.city FROM User u"),
...
@embedded
private City city;

UserBusinessLogic.java:
...
TypedQuery<City> typedQuery = entityManager.createNamedQuery("User.all.cities",User.class);
List<City> names = typedQuery.getResultList();
...

它给出：类型不匹配无法将列表转换为列表。我尝试了两次获取第一个用户，然后在 getResult 上尝试了一个城市，但下面一行出现了相同的错误。

我看到了一些例子，但并没有真正告诉我如何使用正确的代码（仅是 SQL 语法）来获取它。

谢谢你的帮助

杰西

原文

I want to get an list of distinct City's from my User table. I thought the code below would work but gives an error:

User.java
@Entity
@Table(name="user_info")
...
@NamedQuery(name = "User.all.cities", query = "SELECT distinct u.city FROM User u"),
...
@embedded
private City city;

UserBusinessLogic.java:
...
TypedQuery<City> typedQuery = entityManager.createNamedQuery("User.all.cities",User.class);
List<City> names = typedQuery.getResultList();
...

It gives: type mismatch can not convert List to List. I tried two get first user then on getResult a City but same error one line below.

I see some examples but not really tell how to get it with correct code just the SQL syntax.

Thanks for your help

Jess

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衣神在巴黎 2024-12-17 12:13:59

第一件看起来不太好的事情是，您请求 City 对象，但声明获取 User 对象。

你有：

@NamedQuery(name = "User.all.cities", 
            query = "SELECT distinct u.city FROM User u"),

TypedQuery<City> typedQuery = 
                  entityManager.createNamedQuery("User.all.cities", User.class);

它应该在哪里是：

TypedQuery<City> typedQuery = 
                  entityManager.createNamedQuery("User.all.cities", City.class);

The first thing which doesn't look good is that you ask for City objects but declare to get User's.

You have:

@NamedQuery(name = "User.all.cities", 
            query = "SELECT distinct u.city FROM User u"),

TypedQuery<City> typedQuery = 
                  entityManager.createNamedQuery("User.all.cities", User.class);

Where it should be:

TypedQuery<City> typedQuery = 
                  entityManager.createNamedQuery("User.all.cities", City.class);

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緦唸λ蓇 2024-12-17 12:13:59

有效的代码是：

results = getJpaTemplate().execute(new JpaCallback<List<City>>() {
  @Override
  public List<City> doInJpa(EntityManager em) throws PersistenceException {
    TypedQuery<City> query = em.createNamedQuery("User.all.cities", City.class);
    return query.getResultList() ;
   }
}) ;

The code that works is:

results = getJpaTemplate().execute(new JpaCallback<List<City>>() {
  @Override
  public List<City> doInJpa(EntityManager em) throws PersistenceException {
    TypedQuery<City> query = em.createNamedQuery("User.all.cities", City.class);
    return query.getResultList() ;
   }
}) ;

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