REST服务:返回异常的正确方法

Question

我使用Jersey API作为我的REST服务.我的问题是:是否有更优雅的方式以JSON形式返回异常？关心自己创建一个json对象并将其直接附加到响应是否更好？

这是服务中某个方法的简化示例.如您所见,我只使用HashMap,因为该方法可能会抛出异常,在这种情况下,我需要返回有关It的信息.

@Path("/admin")
public class AdminService {

    @Context
    HttpServletRequest request;

    @POST
    @Produces(MediaType.APPLICATION_JSON)
    public Map<Integer, String> createCompany(Company company){

        Map<Integer, String> map = new HashMap<>();
        try{
            AdminFacade adminFacade = (AdminFacade)Utility.getFacade(request);
            adminFacade.createCompany(company);
            map.put(1,"success");
        } catch (ExceptionREST e) {
            map.put(e.getErrorNumber(), e.getMessage());
        } finally {
            return map;
        }
    }
}

Answer 1

您可以创建一个类似下面的类来表示错误,

@JsonPropertyOrder({ "code", "field", "message" })
public class ErrorInfo {

    private String code;

    private String field;

    private String message;

    public String getCode() {
        return code;
    }

    public void setCode(String code) {
        this.code = code;
    }

    public String getField() {
        return field;
    }

    public void setField(String field) {
        this.field = field;
    }

    public String getMessage() {
        return message;
    }

    public void setMessage(String message) {
        this.message = message;
    }

}

你可以创建一个扩展这样的异常的类,

public class InvalidInputException extends RuntimeException {

    private static final long serialVersionUID = -5027121014723838738L;

    private List<ErrorInfo> errors;

    public List<ErrorInfo> getErrors() {
        return this.errors;
    }

    public InvalidInputException(List<ErrorInfo> errors) {
        super();
        this.errors = errors;
    }

    public InvalidInputException(String message, List<ErrorInfo> errors) {
        super(message);
        this.errors = errors;
    }

}

并有一个异常映射器,您可以将List转换为json并返回到具有http状态代码400(错误请求)的用户.

@Provider
public class InvalidInputExceptionMapper implements ExceptionMapper<InvalidInputException> {

    @Override
    @Produces(MediaType.APPLICATION_JSON)
    public Response toResponse(InvalidInputException e) {

        ResponseBuilder rb = Response.status(Status.BAD_REQUEST);

        rb.entity(e.getErrors());
        return rb.build();   
    }
}

Http响应将是,

HTTP/1.1 400 BAD REQUEST
{
"errors": [{
    "error": {
        "code": "100",
        "field": null,
        "message": "Name is required"
    },
    "error": {
        "code": "100",
        "field": null,
        "message": "Age is required"
    }
}]

}