How do I pass the "/" character in a GET request?

Question

I have an api:

http://localhost:8080/mylocalserver/#/reports/{parameter1}/{parameter2}

now the first parameter is parameter1 = "12345 / EF" and the second parameter is parameter2 = "Text"

I am using Spring Rest Controller to create api. When a query is executed with the following parameters above, it shows Request Not Found.

I tried to encode and decode the parameters using URLEncoder.encode, as well as using UriUtils.encode(param, StandardCharsets.UTF_8.name()), but still getting the same error.

Below is the code I tried.

Suppose I have two variables from user input parameter1andparameter2

Now I am creating a URL with the following parameter

private void createApi(String parameter1, String parameter2) {
    try {
      String uri = " http://localhost:8080/mylocalserver/#/reports/"+encode(parameter1)+"/"+encode(parameter2)+" ";

      ServletRequestAttributes requestAttributes =
          (ServletRequestAttributes) RequestContextHolder.getRequestAttributes();
      HttpServletRequest request = requestAttributes.getRequest();
      HttpServletResponse response = requestAttributes.getResponse();
      request.getRequestDispatcher(uri).forward(request, response);
    } catch (Exception e) {
      LOG.error(e.getMessage(), e);
    }
  }

And my coding method:

public String encode(String param) {
    try {
     UriUtils.encode(param, StandardCharsets.UTF_8.name())
      }
    } catch (Exception e) {
      LOG.error(e.getMessage(), e);
    }
    return param;
  }

Note. I am using Tomcat 8.5

Answer 1

:

return.

public String encode(String param) {
    try {
        //missing return at this line in your code
        return UriUtils.encode(param, StandardCharsets.UTF_8.name())
      }
    } catch (Exception e) {
      LOG.error(e.getMessage(), e);
    }
    return param;
  }

BTW .