为什么我要用Spring靴Rest404

Question

我正在使用Spring Boot实现rest服务。实体类在一个单独的包中定义。所以我在A中添加了组件注释pplication.java.

@Configuration
@EnableAutoConfiguration
@ComponentScan("org.mdacc.rists.cghub.model")
@EnableJpaRepositories(basePackages = "org.mdacc.rists.cghub.model") 
public class Application 
{
    public static void main( String[] args )
    {
        SpringApplication.run(Application.class, args);
    }
}

这是我的控制器类：

// SeqController.java
@RestController
public class SeqController {

    @Autowired
    private SeqService seqService;
    @RequestMapping(
            value = "/api/seqs", 
            method = RequestMethod.GET, 
            produces = MediaType.APPLICATION_JSON_VALUE)
    public ResponseEntity<List<SeqTb>> getSeqs() {
        List<SeqTb> seqs = seqService.findAll();

        return new ResponseEntity<List<SeqTb>>(seqs, HttpStatus.OK);
    }
}

我还创建了一个JPA数据存储库，扩展了JPARepository，并在其中添加了自定义查询代码。

// SeqRepository.java
@Repository
public interface SeqRepository extends JpaRepository<SeqTb, Integer> {

    @Override
    public List<SeqTb> findAll();

    @Query("SELECT s FROM SeqTb s where s.analysisId = :analysisId")
    public SeqTb findByAnalysisId(String analysisId);
}

下面是实现服务接口的servicebean类

// SeqServiceBean.java
@Service
public class SeqServiceBean implements SeqService {
    @Autowired
    private SeqRepository seqRepository;

    @Override
    public List<SeqTb> findAll() {
        List<SeqTb> seqs = seqRepository.findAll();
        return seqs;
    }

    public SeqTb findByAnalysisId(String analysisId) {
        SeqTb seq = seqRepository.findByAnalysisId(analysisId);
        return seq;
    }
}

当我启动应用程序并在浏览器中键入以下url时“http://localhost:8080/api/seqs“，我得到404错误。我错过了什么？

编辑#1：我决定取出JPA存储库的东西并将控制器类更改为以下内容：

@RestController
//@RequestMapping("/")
public class SeqController {
    private static BigInteger nextId;
    private static Map<BigInteger, Greeting> greetingMap;

    private static Greeting save(Greeting greeting) {
        if(greetingMap == null) {
            greetingMap = new HashMap<BigInteger, Greeting>();
            nextId = BigInteger.ONE;
        }
        greeting.setId(nextId);
        nextId = nextId.add(BigInteger.ONE);
        greetingMap.put(greeting.getId(), greeting);
        return greeting;
    }

    static {
        Greeting g1 = new Greeting();
        g1.setText("Hello World!");
        save(g1);

        Greeting g2 = new Greeting();
        g1.setText("Hola Mundo!");
        save(g2);

    }
    @RequestMapping(
            value = "/api/greetings", 
            method = RequestMethod.GET, 
            produces = MediaType.APPLICATION_JSON_VALUE)
    public ResponseEntity<Collection<Greeting>> getGreetings() {
        Collection<Greeting> greetings = greetingMap.values();
        return new ResponseEntity<Collection<Greeting>>(greetings, HttpStatus.OK);

    }
}

当我启动应用程序并在浏览器中输入“localhost:8080/api/greetings”时，我仍然得到404。

Answer 1

我将尝试的第一件事是在控制器的类定义上放置@RequestMapping（“/”）。在方法上保持相同的值。

另一件与您的问题无关的事情是，您不需要定义自定义查询。JPA实际上足够智能，只需使用该方法名即可执行您定义的查询。在这里查看findByLastName示例：https://spring.io/guides/gs/accessing-data-jpa/.

Answer 2

如果您的pom中没有spring boot starter web。然后添加xml，原因可能是代码无法映射endpoint。

<dependency>
    <groupId>org.springframework.boot</groupId>
    <artifactId>spring-boot-starter-web</artifactId>
</dependency>

Answer 3

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