Java Hibernate HQL如何多次连接同一张表？

Question

我的表在Java中的结构如下：

class Owner {
    Long id; //primary key
    int age;
    List<Pet> pets; //one-to-many reference
    //...
}
class Pet {
    PetId id;
    Owner owner;
    //...
}
class PetId {
    BigDecimal ownerId;
    String type; //dog, cat, fish, archaeopteryx
}

并在基础SQL Server数据库中：

Owner:
    id numeric(38,0) [pk]
    age numeric(38,0)
    //...
Pet:
    ownerId numeric(38,0) [pk]
    type varchar(30) [pk]
    //...

我正在尝试获取给定age所有具有两种给定types Pets的Owners的列表。

---

我有一个在SQL Server中正常运行的查询：

select distinct o.id from Owner
inner join Pet pet1 on pet1.owner = o.id and pet1.type = @type1
inner join Pet pet2 on pet2.owner = o.id and pet2.type = @type2
where o.age = @age
order by o.id asc

我想将其转换为HQL。 我可能会使用带有此语法的NativeQuery，但我想尽可能避免这种情况。

---

SUB JOIN

我最初的尝试是对本机SQL的几乎直接转换：

List<int> = em.createQuery("select distinct o.id from Owner o " +
" join o.pets pet1 " +
" join o.pets pet2 " +
" where o.age = :age" +
" and pet1.type = :type1 " +
" and pet2.type = :type2 " +
" order by o.id asc ")
.setParameter("age", age)
.setParameter("type1", type1)
.setParameter("type2", type2)
.getResultList();

我的问题是，这会产生org.hibernate.exception.SQLGrammarException ：“无法执行查询”。

---

交叉加入

我想知道同一子数据是否无法多次连接，是否可能需要交叉连接：

List<int> = em.createQuery("select distinct o.id from Owner o, Pet pet1, Pet pet2 " +
" where o.age = :age" +
" and pet1.owner.id = o.id " +
" and pet2.owner.id = o.id " +
" and pet1.type = :type1 " +
" and pet2.type = :type2 " +
" order by o.id asc ")
.setParameter("age", age)
.setParameter("type1", type1)
.setParameter("type2", type2)
.getResultList();

但这会产生相同的错误。

---

子查询

我查看了HQL文档，发现了子查询语法：

List<int> = em.createQuery("select distinct o.id from Owner o " +
" where o.age = :age" +
" and o.id in (select pet1.owner.id " +
"              from Pets pet1 " +
"              where pet1.type = :type1 ) " +
" and o.id in (select pet2.owner.id " +
"              from Pets pet2 " +
"              where pet2.type = :type2 ) " +
" order by o.id asc ")
.setParameter("age", age)
.setParameter("type1", type1)
.setParameter("type2", type2)
.getResultList();

但这又产生了相同的错误。 我在查询中错过了什么或做错了什么，或者使用HQL无法做到？

Answer 1

尝试这个

    List<int> results = em.createQuery("select distinct o.id from Owner o " +
        " where o.age = :age "
        " and o.id in (select pet1.owner.id from Pets pet1 " +
        " where pet1.type = :type1) " +
        " and o.id in (select pet2.owner.id " +
        " from Pets pet2 where pet2.type = :type2) " + 
        " order by o.id asc ")
        .setParameter("age", age)
        .setParameter("type1", type1)
        .setParameter("type2", type2)
        .getResultList();

Answer 2

我发现了我的问题！

由于某些原因，HQL不能轻易在同一查询中允许DISTINCT和ORDER BY 。 我可以使用子查询格式消除对DISTINCT的需要，因为o.id是主键（唯一），所以唯一可以重复的方法是使用JOIN 。

List<int> = em.createQuery("select o.id from Owner o " +
" where o.age = :age" +
" and o.id in (select pet1.owner.id " +
"              from Pets pet1 " +
"              where pet1.type = :type1 ) " +
" and o.id in (select pet2.owner.id " +
"              from Pets pet2 " +
"              where pet2.type = :type2 ) " +
" order by o.id asc ")
.setParameter("age", age)
.setParameter("type1", type1)
.setParameter("type2", type2)
.getResultList();