How to validated rest url in spring boot? - spring

Question

validate Rest URL in spring boot.
Requirement: If I hit the wrong URL then it should throw a custom exception.
ex. Correct URL is "/fulfillment/600747l/send_to_hub" If I hit "/api/600747l/send_to_hub_1" then it should return exception like
"404:- URL not Found.".
Right now it returning "500 : -
{
"timestamp": 1531995246549,
"status": 500,
"error": "Internal Server Error",
"message": "Invalid Request URL.",
"path": "/api/600747l/send_to_hub_1"
}"

Answer 1

you need to write NewClass with annotation #ControllerAdvice which will redirect all exceptions to this NewClass.
example
Your Custom Exception Class:
#Data
#AllArgsConstructor
#EqualsAndHashCode(callSuper = false)
public class IOApiException extends IOException {
private ErrorReason errorReason;
public IOApiException(String message, ErrorReason errorReason) {
super(message);
this.errorReason = errorReason;
}
}
Now the CustomExceptionHandler Class -
#ControllerAdvice
#RestController
public class GlobalExceptionHandler {
Logger logger = LoggerFactory.getLogger(this.getClass());
#ResponseStatus(HttpStatus.UNAUTHORIZED)
#ExceptionHandler(value = IOApiException.class)
public GlobalErrorResponse handleException(IOApiException e) {
logger.error("UNAUTHORIZED: ", e);
return new GlobalErrorResponse("URL Not Found", HttpStatus.UNAUTHORIZED.value(), e.getErrorReason());
}
//this to handle customErrorResponseClasses
public GlobalErrorResponse getErrorResponseFromGenericException(Exception ex) {
if (ex == null) {
return handleException(new Exception("INTERNAL_SERVER_ERROR"));
}
else if (ex instanceof IOApiException) {
return handleException((IOApiException) ex);
}
}
Now Your error response class:
public class GlobalErrorResponse {
private String message;
#JsonIgnore
private int statusCode;
private ErrorReason reason;
}
ErrorReason Class
public enum ErrorReason {
INTERNAL_SERVER_ERROR,
INVALID_REQUEST_PARAMETER,
INVALID_URL
}
add and register one filter who calls the GlobalExceptionHandler in exception case like this
public class ExceptionHandlerFilter implements Filter {
private final GlobalExceptionHandler globalExceptionHandler;
public ExceptionHandlerFilter(GlobalExceptionHandler globalExceptionHandler) {
this.globalExceptionHandler = globalExceptionHandler;
}
#Override
public void init(FilterConfig filterConfig) throws ServletException {
}
#Override
public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException {
try {
chain.doFilter(request, response);
} catch (Exception exception) {
HttpServletResponse httpResponse = (HttpServletResponse) response;
GlobalErrorResponse errorResponse = globalExceptionHandler.getErrorResponseFromGenericException(exception);
httpResponse.setStatus(errorResponse.getStatusCode());
response.getWriter().write(new ObjectMapper().writeValueAsString(errorResponse));
}
}
#Override
public void destroy() {
}
}
Like this you can add as many exceptions you want.. and can handle it manually.

Answer 2

As per your question first of all you need to define a base url(e.g.-/api) so that any url must be handled through your controller.Now after base url as shown /api/600747l/send_to_hub_1 #PathVariable int id. This circumstance is important, because Spring documentation said that if method argument annotated with #PathVariable can’t be casted to specified type (in our case to int), it will be exposed as String. Hence it can cause a TypeMismatchException.
To handle this I will use #ExceptionHandler annotation on #Controller level. Such approach suits for this situation as no one else. I just need to make 2 changes in the Controller:
1.Add MessageSource field
2.Add exception handler method
#Autowired
private MessageSource messageSource;
...
#ExceptionHandler(TypeMismatchException.class)
#ResponseStatus(value=HttpStatus.NOT_FOUND)
#ResponseBody
public ErrorInfo handleTypeMismatchException(HttpServletRequest req, TypeMismatchException ex) {
Locale locale = LocaleContextHolder.getLocale();
String errorMessage = messageSource.getMessage("error.bad.smartphone.id", null, locale);
errorMessage += ex.getValue();
String errorURL = req.getRequestURL().toString();
return new ErrorInfo(errorURL, errorMessage);
}
...