How to set a parameter in a HttpServletRequest?

Question

I am using a javax.servlet.http.HttpServletRequest to implement a web application.

I have no problem to get the parameter of a request using the getParameter method. However I don't know how to set a parameter in my request.

Answer 1

You can't, not using the standard API. HttpServletRequest represent a request received by the server, and so adding new parameters is not a valid option (as far as the API is concerned).

You could in principle implement a subclass of HttpServletRequestWrapper which wraps the original request, and intercepts the getParameter() methods, and pass the wrapped request on when you forward.

If you go this route, you should use a Filter to replace your HttpServletRequest with a HttpServletRequestWrapper:

public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain filterChain) throws IOException, ServletException {     if (servletRequest instanceof HttpServletRequest) {         HttpServletRequest request = (HttpServletRequest) servletRequest;         // Check wether the current request needs to be able to support the body to be read multiple times         if (MULTI_READ_HTTP_METHODS.contains(request.getMethod())) {             // Override current HttpServletRequest with custom implementation             filterChain.doFilter(new HttpServletRequestWrapper(request), servletResponse);             return;         }     }     filterChain.doFilter(servletRequest, servletResponse); }