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Home/ Questions/Q 5993183
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T23:38:19+00:00 2026-05-22T23:38:19+00:00

How to get the base url from the jsp request object? http://localhost:8080/SOMETHING/index.jsp , but

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How to get the base url from the jsp request object?
http://localhost:8080/SOMETHING/index.jsp, but I want the part till index.jsp, how is it possible in jsp?

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  1. Editorial Team
    Editorial Team
    2026-05-22T23:38:19+00:00Added an answer on May 22, 2026 at 11:38 pm

    So, you want the base URL? You can get it in a servlet as follows:

    String url = request.getRequestURL().toString();
    String baseURL = url.substring(0, url.length() - request.getRequestURI().length()) + request.getContextPath() + "/";
    // ...
    

    Or in a JSP, as <base>, with little help of JSTL:

    <%@taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
    <%@taglib prefix="fn" uri="http://java.sun.com/jsp/jstl/functions" %>
    <c:set var="req" value="${pageContext.request}" />
    <c:set var="url">${req.requestURL}</c:set>
    <c:set var="uri" value="${req.requestURI}" />
    ...
    <head>
        <base href="${fn:substring(url, 0, fn:length(url) - fn:length(uri))}${req.contextPath}/" />
    </head>
    

    Note that this does not include the port number when it’s already the default port number, such as 80. The java.net.URL doesn’t take this into account.

    See also:

    • Browser can't access/find relative resources like CSS, images and links when calling a Servlet which forwards to a JSP
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