休眠，插入或更新时不选择 [英] Hibernate, insert or update without select

问题描述

我有一个属于某些类别的产品对象，即经典的多对一关系.

I have a products objects which belongs to certain categories i.e. classical many to one relationship.

@Entity
public class Product{

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    Long id;

    String name;
    Double price;
    @ManyToOne(fetch = FetchType.LAZY)
    Category category;
...
}

@Entity
public class Category implements Identifiable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    private String name;
...
}

我想插入和更新产品而无需预先选择类别.像这样:

I want to insert and update products without preselecting categories. Like this:

Product product = dao.get(productId);
Category category = dao.get(categoryId);
product.setCategory(category);
dao.update(product);

或

Product product = new Product(somename);
Category category = dao.get(categoryId);
product.setCategory(category);
dao.insert(product);

是否可以在不选择类别的情况下进行更新和插入?我不想为此使用HQL或直接查询.

Is it possible to update and insert without category selecting? I don't want to use HQL or direct queries for that.

推荐答案

session.load()专门用于此类情况.以下:

session.load() exists specifically for cases like this. The following:

Category category = session.load(categoryId);
product.setCategory(category);

不会命中数据库.但是，如果没有提供给定ID的类别，它将在稍后阶段(刷新期间或多或少)引发异常.

will not hit the database. It will, however, throw an exception at a later stage (during flush, more or less) if there is no category available with given id.

使用load()比merge()更快，并且没有副作用(级联等)

Using load() is faster than merge() and has no side-effects (cascading, etc...)

这篇关于休眠，插入或更新时不选择的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！