I would grab the URL of the current JSP web page with its settings example: index.jsp? param = 12
Have you any idea? Thank you
Look at the HttpServletRequest Object, which you can access from your JSP in a scriplet (although that's not pretty). It has many methods for getting the URL of the page, including the parameters. Methods of interest will be:
- getQueryString
- getRequestURI
- getRequestURL
Have a play with them.
You can get it from the HttpServletRequest object which is in EL available by ${pageContext.request}
. The part before the ?
is available by getRequestURL() method and the part after the ?
is available by getQueryString() method. So, in a nutshell:
<p>Request URL: ${pageContext.request.requestURL}</p>
<p>Query string: ${pageContext.request.queryString}</p>
<p>Full URL: ${pageContext.request.requestURL}?${pageContext.request.queryString}</p>
If you want to do this using normal Java code, you'd better use a Servlet for this.
String requestURL = request.getRequestURL().toString();
String queryString = request.getQueryString();
if (queryString != null) requestURL += "?" + queryString;
// ...