我得到以下错误:

***************************
APPLICATION FAILED TO START
***************************

Description:

Parameter 0 of method setApplicant in webService.controller.RequestController required a bean of type 'com.service.applicant.Applicant' that could not be found.


Action:

Consider defining a bean of type 'com.service.applicant.Applicant' in your configuration.

我以前从未见过这个错误,但奇怪的是@Autowire不工作。项目结构如下:

申请人接口

public interface Applicant {

    TApplicant findBySSN(String ssn) throws ServletException;

    void deleteByssn(String ssn) throws ServletException;

    void createApplicant(TApplicant tApplicant) throws ServletException;

    void updateApplicant(TApplicant tApplicant) throws ServletException;

    List<TApplicant> getAllApplicants() throws ServletException;
}

ApplicantImpl

@Service
@Transactional
public class ApplicantImpl implements Applicant {

private static Log log = LogFactory.getLog(ApplicantImpl.class);

    private TApplicantRepository applicantRepo;

@Override
    public List<TApplicant> getAllApplicants() throws ServletException {

        List<TApplicant> applicantList = applicantRepo.findAll();

        return applicantList;
    }
}

现在我应该能够自动连线申请人,并能够访问,但在这种情况下,当我在@RestController中调用它时,它是不工作的:

@RestController
public class RequestController extends LoggingAware {

    private Applicant applicant;

    @Autowired
    public void setApplicant(Applicant applicant){
        this.applicant = applicant;
    }

    @RequestMapping(value="/", method = RequestMethod.GET)
    public String helloWorld() {

        try {
            List<TApplicant> applicantList = applicant.getAllApplicants();

            for (TApplicant tApplicant : applicantList){
                System.out.println("Name: "+tApplicant.getIndivName()+" SSN "+tApplicant.getIndSsn());
            }

            return "home";
        }
        catch (ServletException e) {
            e.printStackTrace();
        }

        return "error";
    }

}

------------------------ 更新1 -----------------------

我添加了

@SpringBootApplication
@ComponentScan("module-service")
public class WebServiceApplication extends SpringBootServletInitializer {

    @Override protected SpringApplicationBuilder configure(SpringApplicationBuilder builder) {
        return builder.sources(WebServiceApplication.class);
    }

    public static void main(String[] args) {
        SpringApplication.run(WebServiceApplication.class, args);
    }

}

错误消失了,但什么也没发生。然而,当我在添加@ComponentScan()之前注释掉所有与RestController中的申请人相关的内容时,我能够返回一个字符串的UI,从而意味着我的RestController正在工作,现在它被跳过了。我丑陋的白色标签错误页现在。

--------------------- 更新2 ------------------------------

我添加了它所抱怨的bean的基本包。错误读取:

***************************
APPLICATION FAILED TO START
***************************

Description:

Parameter 0 of method setApplicantRepo in com.service.applicant.ApplicantImpl required a bean of type 'com.delivery.service.request.repository.TApplicantRepository' that could not be found.


Action:

Consider defining a bean of type 'com.delivery.request.request.repository.TApplicantRepository' in your configuration.

我添加了@ComponentScan

@SpringBootApplication
@ComponentScan({"com.delivery.service","com.delivery.request"})
public class WebServiceApplication extends SpringBootServletInitializer {

    @Override protected SpringApplicationBuilder configure(SpringApplicationBuilder builder) {
        return builder.sources(WebServiceApplication.class);
    }

    public static void main(String[] args) {
        SpringApplication.run(WebServiceApplication.class, args);
    }

}

---------------------------- 更新3 ----------------------

添加:

@SpringBootApplication
@ComponentScan("com")
public class WebServiceApplication extends SpringBootServletInitializer {

仍然在抱怨我的ApplicantImpl类@自动连接我的repo TApplicantRepository到它。


当前回答

如果@Service类被标记为抽象,就会发生这种情况。

其他回答

将Springbootapplication(application.java)文件移动到另一个包中解决了我的问题。保持它与控制器和存储库分离。

我也收到了类似的错误:

Consider defining a bean of type 'A_REPOSITORY_INTERFACE' in your configuration.

然后,根据Akashe的解决方案,我将@ enablejparepository添加到我的主类中。在那之后,我收到了以下错误:

Consider defining a bean of type 'entityManagerFactory' in your configuration.

接下来,我浏览了这里所有的回复,谷歌了很多,阅读了很多其他资源,但没有奏效。

最后,我很幸运地在博客/网站(javatute.com)上找到了解决方案。我只是效仿它。

就像许多人在这里建议的那样,我添加了@ComponentScan(“YOUR_BASE_PACKAGE.*”)和@ entitscan(“YOUR_BASE_PACKAGE.*”)到我的主应用程序类中,然后添加了一个配置包并创建了一个JpaConfig类,如:

package YOUR_BASE_PACKAGE.config;

import org.springframework.context.annotation.Configuration;
import org.springframework.data.jpa.repository.config.EnableJpaRepositories;

@Configuration
@EnableJpaRepositories(basePackages = "YOUR_BASE_PACKAGE")
public class JpaConfig {

}

我关注的博客:

考虑在您的配置中定义一个bean类型

这让我想到:

在类路径资源中创建名称为entityManagerFactory的bean时出错:init方法调用失败

最后是:

Hibernate/JPA中使用Spring Boot和Oracle的多对多映射

基本上,当你在“另一个包”中有你的类应用程序时,就会发生这种情况。例如:

com.server
 - Applicacion.class (<--this class have @ComponentScan)
com.server.config
 - MongoConfig.class 
com.server.repository
 - UserRepository

我在Application.class中解决了这个问题

@SpringBootApplication
@ComponentScan ({"com.server", "com.server.config"})
@EnableMongoRepositories ("com.server.repository") // this fix the problem

另一种不那么优雅的方法是:将所有配置类放在同一个包中。

有可能在实现接口之前尝试@autowired接口。

示例解决方案:

    **HomeController.java**
    class HomeController{

      @Autowired
      UserService userService;
    .....
    }
----------------------------------------------------------------------
    **UserService.java** 
    public interface UserService {
        User findByUsername(String username);
    .....
    }
-----------------------------------------------------------------------
     **UserServiceImpl.java**
     @Service
     public class UserServiceImpl implements UserService{

         public User findByUsername(String username) {
           return userDao.findByUsername(username);
         }
        ....
      }

<i>This is not italic</i>, and [this is not a link](https://example.com)

我也遇到过同样的问题。Mongo DB存储库是由Spring引导标识的,但它并没有为扩展Mongo存储库的存储库接口创建Bean。

在我的案例中,问题是maven pom中“spring + mango”的版本规范不正确。我已经改变了神器的组id,这一切都像魔法一样有效。不需要注释,因为spring boot会处理所有事情。

在我解决问题的过程中,我在网上到处搜索解决方案,并意识到这个问题实际上是与项目配置相关的,任何面临这个问题的人都应该首先检查他们的项目设置,并从spring中启用调试,以获得更多关于失败的详细信息,并密切关注在这个过程中的确切位置,创建失败了。