Question

I know this question has been asked often but I'm unable to find a solution. How can I get a generic type class name in a Spring injected repository?

Here it is my base repository

{
    command: ["edit", "destroy"],
    title: "&nbsp;",
    width: 120
}],
editable: "inline"

this is the interface

public interface UserRepository extends JpaRepository<User, Long>, IUserRepository<User>{
   User findByUsername(String username);
}

and finally here it is the implementation

public interface IUserRepository<T> {
   public List<T> findAllValidEtSiteAndStructure();
}

how can I get type.name? Thanks in advance

Answer 1

您的问题的一般答案可以在文档中看到 - ＆gt; “Spring Data存储库的自定义实现”一章中的http://docs.spring.io/spring-data/jpa/docs/current/reference/html/#repositories.custom-implementations

但我认为在你的情况下这不应该是必要的。你应该能够以下面的方式做到这一点。

public interface UserRepository extends JpaRepository<User, Long> {

   User findByUsername(String username);

   List<User> findByStructureAndSiteAndValid(Structure structure, Site site, boolean valid);
}

Answer 2

基本上，由于type erasure，您无法获得泛型类型。

我要做的是向// this means you want to draw a line on the outer edge // 1 is thickness, and you got more options like color, etc. graphics.lineStyle(1); // draw rect graphics.drawRect(x, y, width, height); // draw circle graphics.drawCircle(x, y, radius);添加一个返回相关类型的抽象方法：

UserRepositoryImpl

然后我将为public abstract Class getType();创建特定的实例，在编译时已知该类型。例如：

UserRepositoryImpl

Answer 3

考虑到您正在使用Spring Framework，请使用下面的代码片段，我已经过测试，它运行得很好：

ResolvableType resolvableType = ResolvableType.forClass(UserRepository.class).as(JpaRepository.class);
System.out.println(resolvableType.getGeneric(0));//User
System.out.println(resolvableType.getGeneric(1));//Long

Spring get generic type class

3 个答案: