Jersey ExceptionMapper 不映射异常

Question

我的 Spring Boot + Jersey REST 服务没有按预期工作。

在 UserController 中引发了 EmailExistsException，但我只收到错误 500。一直。而且我的异常没有被记录。

我怀疑异常处理存在一些配置问题，但不知道在哪里进行设置。有什么建议吗？

@POST
@Path("register")
@Produces(MediaType.APPLICATION_JSON)
public Response register(NewUserPayload newUserPayload) throws EmailExistsException, MessagingException

EmailExistsExceptionMapper

@Provider
public class EmailExistsExceptionMapper extends AbstractExceptionMapper       implements
    ExceptionMapper<EmailExistsException>
{
@Override
public Response toResponse(EmailExistsException e)
{
    ResponseEntity re = new ResponseEntity(org.springframework.http.HttpStatus.BAD_REQUEST);

    return this.errorResponse(HttpStatus.BAD_REQUEST_400, re, e);
}
}

AbstractExceptionMapper

@Slf4j
public abstract class AbstractExceptionMapper
{
protected Response errorResponse(int status, ResponseEntity responseEntity, Throwable t)
{
    StringWriter sw = new StringWriter();
    PrintWriter pw = new PrintWriter(sw);
    t.printStackTrace(pw);
    log.error(sw.toString()); // logging stack trace.

    return customizeResponse(status, responseEntity);
}

private Response customizeResponse(int status, ResponseEntity responseEntity)
{
    return Response.status(status).entity(responseEntity).build();
}
}

build.gradle

compile("org.springframework.boot:spring-boot-starter-web") {
    exclude module: 'spring-boot-starter-tomcat'
}
compile "org.springframework.boot:spring-boot-starter-jetty"
compile "org.springframework.boot:spring-boot-starter-security"
compile "org.springframework.boot:spring-boot-starter-aop"
compile "org.springframework.boot:spring-boot-starter-data-jpa"
compile "org.springframework.boot:spring-boot-starter-thymeleaf"
compile "org.springframework.boot:spring-boot-starter-jersey"
compile 'org.springframework.boot:spring-boot-starter-mail'

Answer 1

解决了我的问题的答案：

我已经放了 packages("package.name");这是我的根包名，异常映射器就像一个魅力。

@Component
@ApplicationPath("/api")
public class JerseyConfig extends ResourceConfig
{
public JerseyConfig()
{
    packages("package.name");
    register(UserController.class);
    register(TimezoneController.class);
}
}

Answer 2

您是否已将自定义 ExceptionMapper 配置为 jax-rs 提供程序，并且您确定您的异常被包装到 EmailExistsException 中吗？你可能需要看看这篇文章。

JAX-RS using exception mappers

Answer 3

我遇到了同样的问题

刚才在<init-param>的<param-value>标签中提到了web.xml文件中的根包

然后它开始像魅力一样工作

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>com.two95.restful</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

Answer 4

如果您使用 ExceptionMapper，则必须注册您的异常映射器：

@Component
@Provider
public class ApiApplicationExceptionMapper implements ExceptionMapper<Exception> {

    @Override
    public Response toResponse(Exception exception) {
       ...
    }
}

在 Jersey 配置类中，扫描带有 @Path 注释的 Jersey Endpoint 类和带有 @Provider 注释的自定义异常映射器进行注册：

@Configuration
public class JerseyConfig extends ResourceConfig {

    @Autowired
    ApplicationContext applicationContext;

    @PostConstruct
    public void init() {
        this.registerEndpoints();
        this.registerExceptionMappers();
    }

    private void registerEndpoints() {
        Map<String, Object> beans = applicationContext.getBeansWithAnnotation(Path.class);
        for (Object apiClass : beans.values()) {
            logger.info("Jersey register: " + apiClass.getClass().getName());
            register(apiClass);
        }
    }

    private void registerExceptionMappers() {
        Map<String, Object> beans = applicationContext.getBeansWithAnnotation(Provider.class);
        for (Object exceptionMapper : beans.values()) {
            logger.info("Jersey exception mapper register: " + exceptionMapper.getClass().getName());
            register(exceptionMapper);
        }
    }