Get request URL in JSP which is forwarded by Servl

问题:

How can I get request URL in JSP which is forwarded by Servlet?

If I run following code in JSP,

System.out.println("servlet path= " + request.getServletPath());
System.out.println("request URL= " + request.getRequestURL());
System.out.println("request URI= " + request.getRequestURI());

then I get the server side path to the JSP. But I want to get the URL as you can see in browser's address bar. I can get it in the Servlet that forwards to the JSP, but I want to get it within the JSP.

回答1:

If you use RequestDispatcher.forward() to route the request from controller to the view, then request URI is exposed as a request attribute named javax.servlet.forward.request_uri. So, you can use

request.getAttribute("javax.servlet.forward.request_uri")

or

${requestScope['javax.servlet.forward.request_uri']}

回答2:

Try this instead:

String scheme = req.getScheme();             
String serverName = req.getServerName(); 
int serverPort = req.getServerPort();    
String uri = (String) req.getAttribute("javax.servlet.forward.request_uri");
String prmstr = (String) req.getAttribute("javax.servlet.forward.query_string");
String url = scheme + "://" +serverName + ":" + serverPort + uri + "?" + prmstr;

Note: You can't get HREF anchor from your url. Example, if you have url "toc.html#top" then you can get only "toc.html"

Note: req.getAttribute("javax.servlet.forward.request_uri") work only in JSP. if you run this in controller before JSP then result is null

You can use code for both variant:

public static String getCurrentUrl(HttpServletRequest req) {
    String url = getCurrentUrlWithoutParams(req);
    String prmstr = getCurrentUrlParams(req);
    url += "?" + prmstr;
    return url;
}

public static String getCurrentUrlParams(HttpServletRequest request) {
    return StringUtil.safeString(request.getQueryString());
}

public static String getCurrentUrlWithoutParams(HttpServletRequest request) {
    String uri = (String) request.getAttribute("javax.servlet.forward.request_uri");
    if (uri == null) {
        return request.getRequestURL().toString();
    }
    String scheme = request.getScheme();
    String serverName = request.getServerName();
    int serverPort = request.getServerPort();
    String url = scheme + "://" + serverName + ":" + serverPort + uri;
    return url;
}

回答3:

To get the current path from within the JSP file you can simply do one of the following:

<%= request.getContextPath() %>
<%= request.getRequestURI() %>
<%= request.getRequestURL() %>

回答4:

Try this,

<c:set var="pageUrl" scope="request">
    <c:out value="${pageContext.request.scheme}://${pageContext.request.serverName}"/>
    <c:if test="${pageContext.request.serverPort != '80'}">
        <c:out value=":${pageContext.request.serverPort}"/>
    </c:if>
    <c:out value="${requestScope['javax.servlet.forward.request_uri']}"/>
</c:set>

I would like to put it in my base template and use in whole app whenever i need to.

回答5:

None of these attributes are reliable because per the servlet spec (2.4, 2.5 and 3.0), these attributes are overridden if you include/forward a second time (or if someone calls getNamedDispatcher). I think the only reliable way to get the original request URI/query string is to stick a filter at the beginning of your filter chain in web.xml that sets your own custom request attributes based on request.getRequestURI()/getQueryString() before any forwards/includes take place.

http://www.caucho.com/resin-3.0/webapp/faq.xtp contains an excellent summary of how this works (minus the technical note that a second forward/include messes up your ability to use these attributes).

回答6:

To avoid using scriplets in the jsp, follow the advice of "divideByZero", and use ${pageContext.request.requestURI} This is a better way to go.

回答7:

Also you could use

${pageContext.request.requestURI}