How to get request's URI from WebRequest in Spring?

Question

I am handling REST exceptions using @ControllerAdvice and ResponseEntityExceptionHandler in a spring Rest webservice. So far everything was working fine until I decided to add the URI path(for which exception has occurred) into the BAD_REQUEST response.

@ControllerAdvice
public class RestResponseEntityExceptionHandler extends ResponseEntityExceptionHandler {

@Override
protected ResponseEntity<Object> handleHttpMessageNotReadable(HttpMessageNotReadableException ex,
        HttpHeaders headers, HttpStatus status, WebRequest request) {
    logger.info(request.toString());
    return handleExceptionInternal(ex, errorMessage(HttpStatus.BAD_REQUEST, ex, request), headers, HttpStatus.BAD_REQUEST, request);
}

private ApiError errorMessage(HttpStatus httpStatus, Exception ex, WebRequest request) {
    final String message = ex.getMessage() == null ? ex.getClass().getName() : ex.getMessage();
    final String developerMessage = ex.getCause() == null ? ex.toString() : ex.getCause().getMessage();
    return new ApiError(httpStatus.value(), message, developerMessage, System.currentTimeMillis(), request.getDescription(false));
}

ApiError is just a Pojo class:

public class ApiError {

    private Long timeStamp;
    private int status;
    private String message;
    private String developerMessage;
    private String path;
}

But WebRequest has not given any api to get the path for which the request failed. I tried: request.toString() returns -> ServletWebRequest: uri=/signup;client=0:0:0:0:0:0:0:1
request.getDescription(false) returns -> uri=/signup
getDescription is pretty close to the requirement, but doesn't meet it. Is there any way to get only the uri part?

Answer 1

Found the solution. Casting WebRequest to ServletWebRequest solved the purpose.

((ServletWebRequest)request).getRequest().getRequestURI().toString()

returns the complete path - http://localhost:8080/signup