CODEWITHSUNDEEP
javahttpservlets
flybywire
flybywire

Reputation: 273542

HttpServletRequest to complete URL

I have an HttpServletRequest object.

How do I get the complete and exact URL that caused this call to arrive at my servlet?

Or at least as accurately as possible, as there are perhaps things that can be regenerated (the order of the parameters, perhaps).

Upvotes: 300

Views: 356447

Answers (12)

abishkar bhattarai
abishkar bhattarai

Reputation: 7641

You can use filter .

@Override
public void doFilter(ServletRequest arg0, ServletResponse arg1, FilterChain arg2) throws IOException, ServletException {
        HttpServletRequest test1=    (HttpServletRequest) arg0;
       
     test1.getRequestURL()); it gives  http://localhost:8081/applicationName/menu/index.action
     test1.getRequestURI()); it gives applicationName/menu/index.action
     String pathname = test1.getServletPath()); it gives //menu/index.action
      
  
    if(pathname.equals("//menu/index.action")){ 
        arg2.doFilter(arg0, arg1); // call to urs servlet or frameowrk managed controller method


       // in resposne 
       HttpServletResponse httpResp = (HttpServletResponse) arg1;
       RequestDispatcher rd = arg0.getRequestDispatcher("another.jsp");     
       rd.forward(arg0, arg1);
}

donot forget to put <dispatcher>FORWARD</dispatcher> in filter mapping in web.xml

Upvotes: 1

Ashwin
Ashwin

Reputation: 499

I had an usecase to generate cURL command (that I can use in terminal) from httpServletRequest instance. I created one method like this. You can directly copy-paste the output of this method directly in terminal

private StringBuilder generateCURL(final HttpServletRequest httpServletRequest) {
    final StringBuilder curlCommand = new StringBuilder();
    curlCommand.append("curl ");

    // iterating over headers.
    for (Enumeration<?> e = httpServletRequest.getHeaderNames(); e.hasMoreElements();) {
        String headerName = (String) e.nextElement();
        String headerValue = httpServletRequest.getHeader(headerName);
        // skipping cookies, as we're appending cookies separately.
        if (Objects.equals(headerName, "cookie")) {
            continue;
        }
        if (headerName != null && headerValue != null) {
            curlCommand.append(String.format(" -H \"%s:%s\" ", headerName, headerValue));
        }
    }

    // iterating over cookies.
    final Cookie[] cookieArray = httpServletRequest.getCookies();
    final StringBuilder cookies = new StringBuilder();
    for (Cookie cookie : cookieArray) {
        if (cookie.getName() != null && cookie.getValue() != null) {
            cookies.append(cookie.getName());
            cookies.append('=');
            cookies.append(cookie.getValue());
            cookies.append("; ");
        }
    }
    curlCommand.append(" --cookie \"" + cookies.toString() + "\"");

    // appending request url.
    curlCommand.append(" \"" + httpServletRequest.getRequestURL().toString() + "\"");
    return curlCommand;
}

Upvotes: 0

Jazzschmidt
Jazzschmidt

Reputation: 1105

When the request is being forwarded, e.g. from a reverse proxy, the HttpServletRequest.getRequestURL() method will not return the forwarded url but the local url. When the x-forwarded-* Headers are set, this can be easily handled:

public static String getCurrentUrl(HttpServletRequest request) {
    String forwardedHost = request.getHeader("x-forwarded-host");

    if(forwardedHost == null) {
        return request.getRequestURL().toString();
    }

    String scheme = request.getHeader("x-forwarded-proto");
    String prefix = request.getHeader("x-forwarded-prefix");

    return scheme + "://" + forwardedHost + prefix + request.getRequestURI();
}

This lacks the Query part, but that can be appended as supposed in the other answers. I came here, because I specifically needed that forwarding stuff and can hopefully help someone out with that.

Upvotes: 0

Johannes Stadler
Johannes Stadler

Reputation: 787

You can write a simple one liner with a ternary and if you make use of the builder pattern of the StringBuffer from .getRequestURL():

private String getUrlWithQueryParms(final HttpServletRequest request) { 
    return request.getQueryString() == null ? request.getRequestURL().toString() :
        request.getRequestURL().append("?").append(request.getQueryString()).toString();
}

But that is just syntactic sugar.

Upvotes: 2

Bozho
Bozho

Reputation: 597106

The HttpServletRequest has the following methods:

  • getRequestURL() - returns the part of the full URL before query string separator character ?
  • getQueryString() - returns the part of the full URL after query string separator character ?

So, to get the full URL, just do:

public static String getFullURL(HttpServletRequest request) {
    StringBuilder requestURL = new StringBuilder(request.getRequestURL().toString());
    String queryString = request.getQueryString();

    if (queryString == null) {
        return requestURL.toString();
    } else {
        return requestURL.append('?').append(queryString).toString();
    }
}

Upvotes: 473

ziesemer
ziesemer

Reputation: 28687

Somewhat late to the party, but I included this in my MarkUtils-Web library in WebUtils - Checkstyle-approved and JUnit-tested:

import javax.servlet.http.HttpServletRequest;

public class GetRequestUrl{
    /**
     * <p>A faster replacement for {@link HttpServletRequest#getRequestURL()}
     *  (returns a {@link String} instead of a {@link StringBuffer} - and internally uses a {@link StringBuilder})
     *  that also includes the {@linkplain HttpServletRequest#getQueryString() query string}.</p>
     * <p><a href="https://gist.github.com/ziesemer/700376d8da8c60585438"
     *  >https://gist.github.com/ziesemer/700376d8da8c60585438</a></p>
     * @author Mark A. Ziesemer
     *  <a href="http://www.ziesemer.com.">&lt;www.ziesemer.com&gt;</a>
     */
    public String getRequestUrl(final HttpServletRequest req){
        final String scheme = req.getScheme();
        final int port = req.getServerPort();
        final StringBuilder url = new StringBuilder(256);
        url.append(scheme);
        url.append("://");
        url.append(req.getServerName());
        if(!(("http".equals(scheme) && (port == 0 || port == 80))
                || ("https".equals(scheme) && port == 443))){
            url.append(':');
            url.append(port);
        }
        url.append(req.getRequestURI());
        final String qs = req.getQueryString();
        if(qs != null){
            url.append('?');
            url.append(qs);
        }
        final String result = url.toString();
        return result;
    }
}

Probably the fastest and most robust answer here so far behind Mat Banik's - but even his doesn't account for potential non-standard port configurations with HTTP/HTTPS.

See also:

Upvotes: 0

Peter Szanto
Peter Szanto

Reputation: 7722

In a Spring project you can use 

UriComponentsBuilder.fromHttpRequest(new ServletServerHttpRequest(request)).build().toUriString()

Upvotes: 35

MatBanik
MatBanik

Reputation: 26860

I use this method:

public static String getURL(HttpServletRequest req) {

    String scheme = req.getScheme();             // http
    String serverName = req.getServerName();     // hostname.com
    int serverPort = req.getServerPort();        // 80
    String contextPath = req.getContextPath();   // /mywebapp
    String servletPath = req.getServletPath();   // /servlet/MyServlet
    String pathInfo = req.getPathInfo();         // /a/b;c=123
    String queryString = req.getQueryString();          // d=789

    // Reconstruct original requesting URL
    StringBuilder url = new StringBuilder();
    url.append(scheme).append("://").append(serverName);

    if (serverPort != 80 && serverPort != 443) {
        url.append(":").append(serverPort);
    }

    url.append(contextPath).append(servletPath);

    if (pathInfo != null) {
        url.append(pathInfo);
    }
    if (queryString != null) {
        url.append("?").append(queryString);
    }
    return url.toString();
}

Upvotes: 172

Kishor Prakash
Kishor Prakash

Reputation: 8151

Use the following methods on HttpServletRequest object

java.lang.String getRequestURI() -Returns the part of this request's URL from the protocol name up to the query string in the first line of the HTTP request.

java.lang.StringBuffer getRequestURL() -Reconstructs the URL the client used to make the request.

java.lang.String getQueryString() -Returns the query string that is contained in the request URL after the path.

Upvotes: 1

Vinko Vrsalovic
Vinko Vrsalovic

Reputation: 340241

HttpUtil being deprecated, this is the correct method

StringBuffer url = req.getRequestURL();
String queryString = req.getQueryString();
if (queryString != null) {
    url.append('?');
    url.append(queryString);
}
String requestURL = url.toString();

Upvotes: 7

Teja Kantamneni
Teja Kantamneni

Reputation: 17472

// http://hostname.com/mywebapp/servlet/MyServlet/a/b;c=123?d=789

public static String getUrl(HttpServletRequest req) {
    String reqUrl = req.getRequestURL().toString();
    String queryString = req.getQueryString();   // d=789
    if (queryString != null) {
        reqUrl += "?"+queryString;
    }
    return reqUrl;
}

Upvotes: 28

Michael Borgwardt
Michael Borgwardt

Reputation: 346310

Combining the results of getRequestURL() and getQueryString() should get you the desired result.

Upvotes: 5

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