40
votes

I'm trying to get a correct current URL in JSP in Spring webapp. I'm trying to use the following fragment in the JSP file:

${pageContext.request.requestURL}

The issue is that the returned URL contains prefix and suffix defined by UrlBasedViewResolver. For example the correct URL is:

http://localhost:8080/page

But the returned one is:

http://localhost:8080/WEB-INF/jsp/page.jsp

10
Do you need the absolute URL? Or you just want to build a internal link to another page in your application?jddsantaella
I need the absolute URL.user405935

10 Answers

51
votes

The best way would be to use EL like this:

${requestScope['javax.servlet.forward.request_uri']}
6
votes

Maybe you are looking for something like:

<%= new UrlPathHelper().getOriginatingRequestUri(request) %>

This is not that elegant but solved my problem.

6
votes

In a jsp file:

request.getAttribute("javax.servlet.forward.request_uri")
4
votes

You can make Interceptor and set request attribute e.g.

  request.setAttribute("__SELF",request.getRequestURI);

and in jsp

  <form action="${__SELF}" ></form>   
3
votes

I just found the right answer for your question. The key is using Spring Tags.

<spring:url value="" />

If you put the value attribute empty, Spring will display the mapping URL set in your @RequestMapping.

3
votes

Try this:

<%@ page import="javax.servlet.http.HttpUtils.*" %>
<%= javax.servlet.http.HttpUtils.getRequestURL(request) %> 
3
votes

Anyone who wants to know about other than the reuqest URI, for example a query string, you can check all the names of the variables in the code of RequestDispatcher(of Servlet API 3.1+) interface.

You can get the query string like this:

${requestScope['javax.servlet.forward.query_string']}
0
votes

Which Spring version are you using? I have tested this with Spring 3.1.1.RELEASE, using the following simple application:

Folder structure
-----------------------------------------------------------------------------------

spring-web
    |
     --- src
          |
           --- main
                 |
                  --- webapp
                         |
                          --- page
                         |     |
                         |      --- home.jsp
                         |
                          --- WEB-INF
                               |
                                --- web.xml
                               |
                                --- applicationContext.xml

home.jsp
-----------------------------------------------------------------------------------

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

<html dir="ltr" lang="en" xmlns="http://www.w3.org/1999/xhtml">
    <head>
        <title>Welcome to Spring Web!</title>
    </head>
    <body>
        Page URL: ${pageContext.request.requestURL}
    </body>
</html>

web.xml
-----------------------------------------------------------------------------------

<?xml version="1.0" encoding="UTF-8"?>

<web-app xmlns="http://java.sun.com/xml/ns/javaee" metadata-complete="true" version="2.5" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<display-name>org.example.web</display-name>

<servlet>
    <servlet-name>spring-mvc-dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/webContext.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>spring-mvc-dispatcher</servlet-name>
    <url-pattern>*.htm</url-pattern>
</servlet-mapping>
</web-app>

applicationContext.xml
-----------------------------------------------------------------------------------

<?xml version="1.0" encoding="UTF-8"?>

<beans xmlns="http://www.springframework.org/schema/beans" xmlns:context="http://www.springframework.org/schema/context" xmlns:mvc="http://www.springframework.org/schema/mvc" xmlns:util="http://www.springframework.org/schema/util" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd     http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.1.xsd http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd http://www.springframework.org/schema/util http://www.springframework.org/schema/util/spring-util-3.1.xsd">

<context:annotation-config />
<context:component-scan base-package="org.example" />

<bean id="viewResolver"
    class="org.springframework.web.servlet.view.InternalResourceViewResolver">
    <property name="prefix" value="/page/" />
    <property name="suffix" value=".jsp" />
</bean>

<mvc:annotation-driven />

On accessing http://localhost:8080/spring-web/page/home.jsp, the URL is correctly displayed as http://localhost:8080/spring-web/page/home.jsp.

-1
votes

I used the following in a similar situation. ${currentUrl} can then be Used where needed. Needs core tag library

<c:url value = "" var = "currentUrl" ></c:url>
-2
votes
*<% String myURI = request.getAttribute("javax.servlet.forward.request_uri").toString(); %>
                <% String[] split = myURI.split("/"); %>
                <% System.out.println("My url is-->"+ myURI
                        + "  My url splitter length --->"+split.length
                        +"last value"+split[4]);%>
<%--                <jsp:param name="split[4]" value="split[4]" /> --%>
                <c:set var="orgIdForController" value="<%= split[4] %>" />
                <a type="button" class="btn btn-default btn-xs"
                    href="${pageContext.request.contextPath}/supplier/add/${orgIdForController}">
                    <span class="glyphicon glyphicon-plus" aria-hidden="true"></span>
                    Add
                </a>
 - List item*