java – Tapestry 5和具有相同接口的Spring bean

我在Tapestry 5和
Spring集成方面遇到了问题.如果我有一个实现相同接口的多个bean并尝试使用@Inject注释注入它们,则会出现问题.当然我得到了一个例外.

我找到了一个tutorial,说在那种情况下我也必须使用@Service注释,但现在我得到了

org.apache.tapestry5.internal.services.TransformationException
Error obtaining injected value for field 
com.foo.pages.Foo.testService: Service 
id 'someServiceIDeclaredInSpringContextFile' is not defined by any module...

无论如何,问题是:如何将两个不同的Spring bean(实现相同的接口)注入Tapestry 5页面?

最佳答案 我解决了这个问题.

首先我做了一个新的注释

public @interface Bean {
    String value();
}

我使用这个,无论我有多个bean实现相同的接口

@Inject
@Bean("springBeanName")
Service foo;

然后我改变了org.apache.tapestry5.internal.spring.SpringModuleDef

private ContributionDef createContributionToMasterObjectProvider() {
  ....
  public void contribute(ModuleBuilderSource moduleSource, 
                ServiceResources resources,
                OrderedConfiguration configuration) {
    ....
    switch (beanMap.size()) {
           case 0:
             return null;
           case 1:
             Object bean = beanMap.values().iterator().next();
             return objectType.cast(bean);
           default:
             Bean annotation = annotationProvider.getAnnotation(Bean.class);
             Object springBean = null;
             String beanName = null;

             if (annotation != null) {
               beanName = annotation.value();
               springBean = beanMap.get(beanName);
             } else {
               String message = String.format(
                 "Spring context contains %d beans assignable to type %s: %s.",
                 beanMap.size(),
                 ClassFabUtils.toJavaClassName(objectType),
                 InternalUtils.joinSorted(beanMap.keySet()));
               throw new IllegalArgumentException(message);
             }
             if (springBean != null) {
               return objectType.cast(springBean);
             } else {
               String message = String.format(
                 "Bean [%s] of type %s doesn't exists. Available beans: %s",
                 beanName, ClassFabUtils.toJavaClassName(objectType),
                 InternalUtils.joinSorted(beanMap.keySet()));
               throw new IllegalArgumentException(message);
             }
           }
         }
       };
点赞