Question

将JpaRepository正确注入数据提供程序时遇到问题。

我将@EnableJpaRepositories注释放在包中的EpisodieApiApplication.class中：

package pl.hypeapp.episodie;

@Configuration
@EnableAutoConfiguration
@ComponentScan(basePackages = "pl.hypeapp.episodie.configuration")
@EnableAdminServer
@EnableJpaRepositories
public class EpisodieApiApplication extends SpringBootServletInitializer {

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

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

位于pacakage的

DatabaseDataProviderConfiguration.class

package pl.hypeapp.episodie.configuration.dataprovider;

@Configuration
public class DatabaseDataProviderConfiguration {

   @Bean
   public TvShowDatabaseProvider tvShowDatabaseProvider(TvShowRepository tvShowRepository) {
   return new TvShowDatabaseProvider(tvShowRepository);
   }
}

位于包

中的

TvShowRepository.class

package pl.hypeapp.episodie.dataproviders.database;

public interface TvShowRepository extends JpaRepository<TvShow, String> {
}

模块应用程序/ dataproviders / database中的我的databaseprovider DatabaseDataProvider.class

public class TvShowDatabaseProvider implements DoesTvShowExist, GetTvShowFromDatabase, InsertTvShowToDatabase {

    private final TvShowRepository tvShowRepository;

    public TvShowDatabaseProvider(TvShowRepository tvShowRepository) {
        this.tvShowRepository = tvShowRepository;
    }
}

位于包中的

TvShow.class：

package pl.hypeapp.core.entity.tvmaze;

@JsonIgnoreProperties(ignoreUnknown = true)
@Data
@Entity
public class TvShow {

    @JsonProperty("id")
    public Integer id;
}

但我收到了这个错误：

在类路径资源[pl / hypeapp / episodie / configuration / dataprovider / DatabaseDataProviderConfiguration.class]中定义名称为'tvShowDatabaseProvider'的bean时出错：通过方法'tvShowDatabaseProvider'参数0表示不满意的依赖关系;嵌套异常是org.springframework.beans.factory.BeanCreationException：创建名为'tvShowRepository'的bean时出错：init方法的调用失败;嵌套异常是java.lang.IllegalArgumentException：不是托管类型：class pl.hypeapp.core.entity.tvmaze.TvShow

github处的整个代码，但其不完整，可能由于缺少类而无效

Answer 1

您正在使用构造函数注入来注入TvShowRepository bean，@Autowired的构造函数缺少TvShowDatabaseProvider，因此您需要添加它，如下所示：

public class TvShowDatabaseProvider implements 
             DoesTvShowExist, GetTvShowFromDatabase, 
             InsertTvShowToDatabase {

    private final TvShowRepository tvShowRepository;

    @Autowired //add this
    public TvShowDatabaseProvider(TvShowRepository tvShowRepository) {
        this.tvShowRepository = tvShowRepository;
    }
}

Answer 2

您没有扫描您的实体。您可以使用"pl.hypeapp"作为@EnableJpaRepository的basePackages或使用@EntityScan：

@Configuration
@EnableAutoConfiguration
@ComponentScan(basePackages = "pl.hypeapp.episodie.configuration")
@EnableAdminServer
@EnableJpaRepositories(basePackages = {"pl.hypeapp.episodie.dataproviders"})
@EntityScan(basePackages = {"pl.hypeapp.core.entity"})
public class EpisodieApiApplication extends SpringBootServletInitializer {

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

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

注入JpaRepository：创建名为

2 个答案: