java EntityManager 通过 joinColumn 查询
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EntityManager query by joinColumn
提问by kasavbere
I have a Loginentity and a Customerentity. Login.usernameis a foreign key in the customer table. Hence the following line in the Java CustomerPOJO
我有一个Login实体和一个Customer实体。Login.username是客户表中的外键。因此，Java CustomerPOJO 中的以下行
@OneToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "username", nullable = false)
private Login login;
My question is this: Is there a simple way to query the customertable using username? Or must I first fetch loginby usernameand then customerby login?
我的问题是：有没有一种简单的方法可以customer使用username? 还是必须我第一次取login的username，然后customer用login？
Here is the JPA criteria query. And, yes, I would prefer to use criteria query.
这是 JPA 条件查询。而且，是的，我更喜欢使用标准查询。
public Customer getCustomerByUsername(String username) throws EntityNotFoundException {
CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Customer> criteriaQuery = criteriaBuilder.createQuery(Customer.class);
Root<Customer> root = criteriaQuery.from(Customer.class);
Path<String> path = root.<String>get("username");
criteriaQuery.where(criteriaBuilder.equal(path, username));
return entityManager.createQuery(criteriaQuery).getSingleResult();
}
The line Path<String> path = root.<String>get("username")is throwing an exception saying that username ... is not present.
该行Path<String> path = root.<String>get("username")抛出异常说username ... is not present.
回答by kasavbere
The correct solution with JPQL is
JPQL 的正确解决方案是
Query q = entityManager.createQuery("SELECT c FROM Customer c WHERE c.login.username = :username");
q.setParameter("username", username);
return (Customer) q.getSingleResult();
回答by ppeterka
Query q = entityManager.createQuery("SELECT c FROM Customer c JOIN Login l ON c.login=l WHERE l.username = :username");
q.setParameter("username",username);
List<Customer> customerList = q.getResultList();
There are a few tricks to keep in mind: this all is about the objects, and not the underlying DB. So the names represent classes, instances, and fields, and they are all CaseSensitive... The error you got meant that your Customer class didn't have a userName field - which is true, since Login has that... Sadly, I can't test it, but the logic and intentions should be clear.
有一些技巧需要牢记：这一切都与对象有关，而不是底层数据库。所以名称代表类、实例和字段，它们都是区分大小写的......你得到的错误意味着你的 Customer 类没有 userName 字段 - 这是真的，因为 Login 有......可悲的是，我无法测试，但逻辑和意图应该很清楚。
