标准hibernate加入

Question

我需要做一个返回“b”对象列表的标准。我有以下课程：

---
A
---
private BigDecimal id;
private String name;
private B b;

---
B
---
private BigDecimal id;
private String name;

我这样做：

Criteria criteria = getSession().createCriteria(A.class,"a").createCriteria("b");
return criteria.list();

我总是得到一个List <A>，我不知道如何获得List <B>。这有可能吗？

编辑：使用Java 1.7

Answer 1

如果您不希望A和B之间存在双向关系，请使用别名，然后仅收集b属性。

List<A> aList = getSession().createCriteria(A.class, "a")
        .createAlias("a.b", "b")
        .add(Restrictions.eq("a.name", "A name")
        .add(Restrictions.eq("b.name", "B name")
        .list();
List<B> bList = new ArrayList<>(aList.size());
for (A a : aList) {
    bList.add(a.getB());
}

Answer 2

您可以使用嵌套的Criteria导航到关联，如下所示：

List<B> result = session.createCriteria( A.class )
     .add( Restrictions.like("name","he%") )
     .createCriteria( "b" )
         .add( Restrictions.like("name", "he%") )
     .list();

我的“第二个”答案不适用于Criteria，而是CriteriaBuilder，我不推荐用于小用例，但是用于需要根据来自用户或其他应用程序的不断变化的条件动态创建查询的用例。

final CriteriaBuilder queryBuilder = getCriteriaBuilder(); // Retrieve your CriteriaBuilder here. I inject mine over CDI for example...
final CriteriaQuery<B> query = queryBuilder.createQuery( B.class ); // Type the query to it's expected end result type.

final Root<A> queryRoot = query.from( A.class ); // Select the root of the query
final Join<Object, Object> BJoin= queryRoot.join("b", JoinType.LEFT ); // "b" is the field name to use for the mapping between the root table to the joined table. In this case a.b
// The above equals "left join b on b.id = a.b.id "

// Perform a select with the Class resulting from the select and what wants to be selected. 
// It is also possible to select only a field of a table but in our case we want the whole table of B to be selected.
final Selection<B> select = queryBuilder.construct( B.class, BJoin ); 
query.select(select); // add the select to the query

// We need to remember the ParameterExpression in order to fill the where condition.
// This acts as a typed(!) blank to later fill with the condition we want to match
final ParameterExpression<String> bName = queryBuilder.parameter(String.class);

// Define the where condition using the Path<T> you retrieve from Root or Join objects.
// This will make hibernate build the condition for the correct table like b.name
final Predicate bPredicate = queryBuilder.equal( bJoin.get("name"),bName );

query.where(bPredicate); // add the where expression to the query.
// The above equals something like "where b.name = ?"

// Compile the built query to a TypedQuery
// The EntitiyManager is also injected over CDI in my usual case.
final TypedQuery<B> builtQuery = javax.persistence.EntityManager.createQuery(query);
builtQuery.setParameter(bName,"test"); // Fill in the ? of the where condition.

final List<B> resultList = builtQuery.getResultList();

这在开始时看起来非常沉重和复杂，但是可以非常动态地使用它，因为您可以从代码段中提取多个方法并在条件，顺序，分组等处添加多个方法。

Answer 3

最后我这样做了：

public List<B> findXXXX(A a) {

    // A
    DetachedCriteria aCriteria = DetachedCriteria.forClass(A.class, "a");
    aCriteria.add(Restrictions.eq("a.name", "some name"));
    aCriteria.setProjection(Projections.property("a.key"));

    // B
    Criteria bcriteria = getSession().createCriteria(B.class, "b");
    bcriteria.add(Property.forName("a.key").in(aCriteria));

    return bcriteria.list();
}

子查询因为将在数据库中占用很少的记录。谢谢你的帮助！