Question

我正在编写一个在Hibernate中有多个连接的查询。现在我需要查询使用某个项目的所有模型和相关产品。

我创建了hql查询，它看起来像：

"FROM Model m inner join fetch m.products p inner join p.itemsInProducts iip inner join iip.item i where i.number = '" + itemID + "' and m.active = '1' and p.active = '1' and iip.active = '1'"

它给了我一些结果。但是我有多个Model对象是相同但具有不同的产品。有没有办法组合所有Model对象并将Product放在同一个Model中？

我在Model类中的变量如下所示：

public class Model implements java.io.Serializable {

    private long id;
    private ItemClass itemClass;
    private OrderProcessorCode orderProcessorCode;
    private String name;
    private boolean active;
    private long companyId;
    @JsonManagedReference
    private Set products = new HashSet(0);
    private Set itemsConsumedPerModels = new HashSet(0);
    private Set itemsInModels = new HashSet(0);

产品：

public class Product implements java.io.Serializable {

    private long id;
    private ItemClass itemClass;
    private ItemType itemType;
    private OrderProcessorCode orderProcessorCode;
    private String number;
    private String description;
    private boolean baseProduct;
    private boolean otmProduct;
    private boolean active;
    private long companyId;
    private Set itemsInProducts = new HashSet(0);
    @JsonBackReference(value = "product-models")
    private Set models = new HashSet(0);
    private Set optionsPerProductsForProductId = new HashSet(0);
    private Set optionsPerProductsForOptionId = new HashSet(0);
    private Set productionOrders = new HashSet(0);

Answer 1

稍微更改了我的查询和代码。现在，所有查询过的产品都会自动添加到模型中。

Query query = session.createQuery("FROM Model m inner join fetch m.products p inner join p.itemsInProducts iip inner join iip.item i where i.number = '" + itemID +"'  and m.active = '1' and p.active = '1' and iip.active = '1'");

        ArrayList<Object[]> items = (ArrayList<Object[]>) query.list();

        session.close();

        for (Object[] objects : items) {
            Model tempModel = (Model) objects[0];

            if(!allmodels.contains(tempModel))
            {
                allmodels.add(tempModel);
            }
        }

在Hibernate中将多个结果组合在一起

1 个答案: