android - 如何使用 greenDao 获得递归的多对多关系

Question

我想获得类别的递归数据结构。

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OOM结构:

一个类别应该管理一个列表或一组(子)类别。

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ROM结构:

ROM的物理结构:

category (_id_, title)
overAndSubCategories (ocId, scId)

overAndSubCategories 的两个属性都是外键，并引用 category 的 id。

category的标题是唯一的，可以是主键，但是greenDao目前不支持字符串作为主键。所以我按照记录添加了一个 id。

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GreenDao 不支持多对多关系。我能否实现解决此结构问题的等效方案？

*两张图片均由 yEd 提供支持

Answer 1

我不确定这个解决方案是正确的还是最好的，因为我不是 100% 知道我在做什么，但它确实有效。

方案生成器:

final Entity category = schema.addEntity("Category");
category.addIdProperty();
category.addStringProperty("Name").notNull().unique().index();

final Entity overSubCategory = schema.addEntity("OverSubCategory");
// no own id! we do not want to load objects of this type
final Property fkOverCategory = overSubCategory.addLongProperty("IdO").notNull().getProperty();
final Property fkSubCategory = overSubCategory.addLongProperty("IdS").notNull().getProperty();
overSubCategory.addToOne(category, fkOverCategory, "OverCategory");
overSubCategory.addToOne(category, fkSubCategory, "SubCategory");

该方案的使用:

/* # write # */

final CategoryDao categoryDao = session.getCategoryDao();
final OverSubCategoryDao overSubCategory = session.getOverSubCategoryDao();

// overcategories
final Category oC1 = new Category(1l, "food and drink");
final Category oC2 = new Category(2l, "party");
// subcategories
final Category sC1 = new Category(3l, "junkfood");
final Category sC2 = new Category(4l, "restaurante");
final Category sC3 = new Category(5l, "pub");

categoryDao.insert(oC1);
categoryDao.insert(oC2);
categoryDao.insert(sC1);
categoryDao.insert(sC2);
categoryDao.insert(sC3);

overSubCategory.insert( new OverSubCategory(oC1.getId(), sC1.getId()) );
overSubCategory.insert( new OverSubCategory(oC1.getId(), sC2.getId()) );
overSubCategory.insert( new OverSubCategory(oC1.getId(), sC3.getId()) );
overSubCategory.insert( new OverSubCategory(oC2.getId(), sC3.getId()) );

/* # read # */

final QueryBuilder<Category> cqb = categoryDao.queryBuilder();
// all categories that are subcategories and ...
final Join joinOverSubCategories = cqb.join(OverSubCategory.class, OverSubCategoryDao.Properties.IdS);
// ... whose overcategories ...
final Join joinSubCategories = cqb.join(joinOverSubCategories, OverSubCategoryDao.Properties.IdO, Category.class, CategoryDao.Properties.Id);
// ... named "food and drink".
joinSubCategories.where(CategoryDao.Properties.Name.eq("food and drink"));

final List<Category> someSubcategories = cqb.list();

两个连接加上一个 where 条件似乎有点过分了。需要更好的解决方案。