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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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Each time a client requests a page, the JSP engine creates a new object to represent that request. The request object provides methods to get HTTP header information including form data, cookies, HTTP methods, etc.
The JSP request is an implicit object of type HttpServletRequest i.e. created for each jsp request by the web container. It can be used to get request information such as parameter, header information, remote address, server name, server port, content type, character encoding etc.
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.
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