Question

我有以下情况3表，人，工作，类别。人有工作，工作有一个类别。如何从同一类别中获取人员记录。

 public List<Person> findPplByCategory(Category category) {
    javax.persistence.criteria.CriteriaBuilder cb = getEntityManager().getCriteriaBuilder();
    javax.persistence.criteria.CriteriaQuery cq = getEntityManager().getCriteriaBuilder().createQuery();
    javax.persistence.criteria.Root<Person> e = cq.from(Person.class);
    javax.persistence.criteria.Root<Job> a = cq.from(Job.class);
    //...not sure how to create the query here..
}

Answer 1

使用JPQL查询很容易做到

String query = "select P from Person P join P.job J join J.category C where C = :cat"

List<Person> = entitiyManager.createQuery(query, Person.class).setParameter("cat", myCategory).getResultList();

一些假设：

一个人有一份工作
一个工作有一个类别
人员的工作字段名为job
作业的类别字段名为category

Answer 2

尚未测试，但可能类似

    CriteriaBuilder cb = entityManager.getCriteriaBuilder();
    CriteriaQuery<Person> q = cb.createQuery(Person.class);
    Root<Person> person = q.from(Person.class);
    Join<Person,Job> job = person.join(Person_.job);

    Predicate predicate = cb.conjunction();
    Predicate p = cb.equal(job.get(Job_.category), cat);
    predicate = cb.and(p);

    q.where(predicate);
    q.select(person);

    TypedQuery<Person> tq = entityManager.createQuery(q);
    List<Person> all = tq.getResultList();

    return all;

Answer 3

    javax.persistence.criteria.CriteriaBuilder cb = getEntityManager().getCriteriaBuilder();
    javax.persistence.criteria.CriteriaQuery cq = getEntityManager().getCriteriaBuilder().createQuery();
    javax.persistence.criteria.Root<Person> e = cq.from(perdon.class);
    cq.where(cb.equal(e.get("idCompany").get("idCategory"), cat));
    cq.orderBy(cb.desc(e.get("idPerson")));
    return getEntityManager().createQuery(cq).getResultList();

Answer 4

我的版本是：

CriteriaBuilder cb = getEntityManager().getCriteriaBuilder();
CriteriaQuery<Person> cq = cb.createQuery(Person.class);
Root<Person> root = cq.from(Person.class);
Join<Person,Job> job = root.join(Person_.job);
Join<Job,Category> category =car.join(Car_.category).where(cb.equal(job.get(Job_.category).get(Category_.name),"your category"));
cq.select(root);
TypedQuery<Person> query = getEntityManager().createQuery(cq);
return query.getResultList();

在这里，Person one-to-one为Job，后者为one-to-one Category :)您将收到所有人指定的工作类别。

JPA如何构建连接条件查询

4 个答案: