Issue
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?
Solution
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:
Answered By - BalusC
Answer Checked By - Senaida (JavaFixing Volunteer)
