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Question

Spring中有一种机制可以为通用bean的构建自动化提供一种方法吗？

例如，如果我有一个类似的类定义：

class Foo<T> {
    private final T type;
    ...
}

和依赖性，如：

@Autowired
private Foo<String> foo;

我想在Spring中使用某种机制，该机制以某种形式从依赖项定义（在上面的示例中为String）提供T，并提供一种自动创建实例的方式？

Answer 1

通过在您的Spring配置中定义它：

@Bean
public Foo<String> foo() {
    return new Foo<>("bar");
}

或代替值，您可以指定类型：

@Bean
public Foo<String> foo() {
    return new Foo<>(String.class);
}

如果使用Spring Boot，则可以添加@ConditionalOnProperty或@ConditionalOnBean来动态化bean实例化。

（更新）如果要避免声明，则应扩展DefaultListableBeanFactory和ApplicationContext 。 这是Spring Boot的工作示例：

@Controller
@EnableAutoConfiguration
public class BeanFactoryTest {
    public static void main(String[] args) throws Exception {
        SpringApplication app = new SpringApplication(BeanFactoryTest.class);
        app.setApplicationContextClass(CustomAppContext.class);
        app.run(args);
    }

    @Autowired
    private Foo<String> foo1;

    @Autowired
    private Foo<String> foo2;

    @Autowired
    private Foo<Integer> foo3;

    @PostConstruct
    public void initialize() {
        System.out.println(foo1); // prints BeanFactoryTest$Foo@344b8190
        System.out.println(foo2); // prints BeanFactoryTest$Foo@344b8190
        System.out.println(foo3); // prints BeanFactoryTest$Foo@5b69d40d
    }

    public static class CustomAppContext extends AnnotationConfigApplicationContext {
        public CustomAppContext() {
            super(new CustomBeanFactory());
        }
    }

    public static class CustomBeanFactory extends DefaultListableBeanFactory {
        @Override
        protected Map<String, Object> findAutowireCandidates(String beanName, Class<?> requiredType, DependencyDescriptor descriptor) {
            Map<String, Object> map = super.findAutowireCandidates(beanName, requiredType, descriptor);

            if (Foo.class.isAssignableFrom(requiredType)) {
                ResolvableType type = ResolvableType.forField(descriptor.getField());
                ResolvableType genericType = type.getGeneric(0);
                Class<?> genericTypeRaw = genericType.getRawClass();

                boolean hasInstance =
                        map.values()
                            .parallelStream()
                            .map(Foo.class::cast)
                            .map(Foo::getType)
                            .filter(genericTypeRaw::isAssignableFrom)
                            .findFirst()
                            .isPresent();

                if (!hasInstance) {
                    super.registerResolvableDependency(requiredType, new Foo<>(genericTypeRaw));

                    map = super.findAutowireCandidates(beanName, requiredType, descriptor);
                }
            }

            return map;
        }
    }

    public static class Foo<T> {
        private final Class<T> type;

        public Foo(Class<T> type) {
            this.type = type;
        }

        public Class<T> getType() {
            return type;
        }
    }
}

Answer 2

看来你可以使用ResolvableType在春季4 http://docs.spring.io/spring/docs/4.0.0.RC2/javadoc-api/org/springframework/core/ResolvableType.html还有一个细节样品https://spring.io/blog/2013/12/03/spring-framework-4-0-and-java-generics

希望对您有所帮助。

Answer 3

这个问题有点老了，但是没有包含示例的答案，所以...

使用ResolvableType ：

public <E> Foo<E> produceFoo(final InjectionPoint ip) {
    ResolvableType resolved = ResolvableType.forField(ip.getField());
    Foo<E> fooInstance = new FooImpl<>();

    Class<E> parameterClass = (Class<E>) resolved.getGeneric(0).resolve();
    fooInstance.doSomethingWithParametrized(parameterClass);

    return fooInstance;
}

InjectionPoint来自：import org.springframework.beans.factory.InjectionPoint;