Question

我有三个模型的hibernate数据库：

Article - which has is part of multiple categories (EDIT).
Category - which contains articles, and is part of an App
App - which has different categories

我想选择所有具有特定应用类别的文章。所以我想创建类似的东西：

find("ANY categories.app = ?", app).fetch();

这样的东西对我来说对CoreData有用，但显然没有用JPA，我似乎无法找到如何做到这一点。

编辑：澄清：在文章中：

    @ManyToMany
    public List<Category> categories;

分类中的

：

    @ManyToOne
    public App app;

    @ManyToMany(mappedBy = "categories")
    public List<Article> articles;

在App中：

    @OneToMany(mappedBy = "app")
    public List<Category> categories;

Answer 1

select a from Article a where a.category.app = ?

阅读http://docs.jboss.org/hibernate/core/3.6/reference/en-US/html_single/#queryhql

使用JPQL在模型中选择的最简单方法

1 个答案: