<S 扩展 T> 是什么意思保存(S实体)；在 Spring 存储库中?

问题描述

在 Spring Data 项目中，CrudRepository 为正在管理的实体类提供复杂的 CRUD 功能.

In Spring Data project the CrudRepository provides sophisticated CRUD functionality for the entity class that is being managed.

public interface CrudRepository<T, ID extends Serializable>
extends Repository<T, ID> {

    <S extends T> S save(S entity);

    T findOne(ID primaryKey);

    Iterable<T> findAll();

    Long count();

    void delete(T entity);

    boolean exists(ID primaryKey);

    // … more functionality omitted.
}

一般来说，我知道S extends T"是什么意思，即save操作的返回类型S必须是T的子类型.为什么要加上这样的约束?我认为这样做会很好:

In general, I know what "S extends T" means, i.e. that S, the return type of save operation, must be subtype of T. Why is it necessary to add such as constraint? I think that would be fine doing something like this:

T save (T entity);

如

void delete(T entity);

我在谷歌上搜索以寻求更多帮助，我发现了关于 stackoverflow 本身的以下问题，但对我来说并不是很清楚:

I've googled to find more help and I've figured out following question on stackoverflow itself but it isn't really clear for me:

Spring CrudRepository:为什么要发明一个新的泛型S型

谢谢.

推荐答案

如果你想拥有它

T save (T entity);

那么您可以将结果分配给的唯一变量必须是 T 类型.

Then the only variable you could assign the result to would have to be of type T.

所以，如果你有一个 CrudRepository<Animal,AnimalID>存储库，你有

So, if you have a CrudRepository<Animal,AnimalID> repository, and you have

Dog dog = getDog();
Dog savedDog = repository.save(dog);

你会得到一个编译错误 - 你不能将结果分配给 Dog，因为它必须是 T 类型，在这种情况下，动物.

You'd get a compile error - you can't assign the result to Dog, as it has to be of type T, in this case, Animal.

您需要检查返回的值是否确实是 Dog 类型，如果是，则将其转换为 Dog 以将其放入 savedDog.