Question

我正在尝试为 Alfresco SDK 应用创建测试环境。我需要测试由我创建的服务，该服务具有 ServiceRegistry 作为属性，由Spring通过构造函数注入。如果应用程序正在运行，该服务运行良好。

对于bean声明，我使用注释：

@Service(value = "customFormService")
public class CustomFormService implements FormService<String> {
  private final ServiceRegistry serviceRegistry;

  @Autowired
  public CustomFormService(ServiceRegistry serviceRegistry) {
    this.serviceRegistry = serviceRegistry;
  }

在服务应用程序上下文中：

/amp/config/module/sampleapp/context/service-context.xml

我声明<context:component-scan base-package="com.base"/>启用带注释的组件扫描

问题是当我尝试创建这样的单元测试时：

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(
  locations = {"classpath*:config/alfresco/module/sampleapp/context/service-context.xml"}
)
public class DocumentManagementTest {
  @Autowired
  @Qualifier(value = "customFormService")
  FormService<String> customFormService;

  @Test
  public void testWiring() {
    Assert.assertNotNull(customFormService);
  }
}

我得到一个例外：

Caused by: org.springframework.beans.factory.NoSuchBeanDefinitionException: No qualifying bean of type [com.base.form.service.FormService] found for dependency: expected at least 1 bean which qualifies as autowire candidate for this dependency. Dependency annotations: {@org.springframework.beans.factory.annotation.Autowired(required=true), @org.springframework.beans.factory.annotation.Qualifier(value=customFormService}

当应用程序运行时，注入工作正常并且在Junit测试用例中失败，这非常令人困惑。

Answer 1

如果您想在课程中注入 ServiceRegistry ，则需要添加属性

 <bean id="reject-content" class="com.repo.RejectContent"
        parent="action-executer">
        <property name="serviceRegistry">
            <ref bean="ServiceRegistry" />
        </property>


    </bean>

像你上面的bean一样。 com.repo.RejectContent 在RejectContent类中，您需要为服务注册表创建setter，如下所示：

private ServiceRegistry serviceRegistry;

public void setServiceRegistry(ServiceRegistry serviceRegistry) {
    this.serviceRegistry = serviceRegistry;
}

之后，您可以使用 serviceRegistry 。

Answer 2

import org.springframework.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;
...

ApplicationContext applicationContext = new ClassPathXmlApplicationContext(new String[] { "classpath:alfresco/application-context.xml" });
ServiceRegistry serviceRegistry = (ServiceRegistry) applicationContext.getBean(ServiceRegistry.SERVICE_REGISTRY);
...

Alfresco ServiceRegistry无法注入Junit

2 个答案: