How to show the requested URL in a JSP error page?

Question

In the error page I would like to display the URL what the user requested.

In my web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
        xmlns="http://java.sun.com/xml/ns/javaee"
        xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
        xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
                            http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
        id="WebApp_ID"
        version="3.0">

    <display-name>MyStuff</display-name>

    <error-page>
        <error-code>404</error-code>
        <location>/WEB-INF/error-404.jsp</location>
    </error-page>

</web-app>

This will forward to error-404.jsp, and here is the content of that file:

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
    "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
    <title>Page Not found</title>
</head>
<body>
    <p align="center">
    <%
        out.println("Requested resource: " + request.getRequestURL()+ " not found");
    %>
</body>
</html>

The problem it is the request.getRequestURL() need to be changed, but don't know the keyword for what to search.

When I start the browser for http://localhost:8080/MyStuff then I get the following error:

Requested resource: http://localhost:8080/MyStuff/WEB-INF/error-404.jsp not found

How to solve this?

If you looking to catch the error request uri in servlet, then this might help. `(String) req.getAttribute("javax.servlet.error.request_uri");` — Sorter, Dec 08 '13 at 18:17

score 8 · Accepted Answer · edited May 23 '17 at 11:58

Here is a simple example of JSP error page that shows the error code and the URL of the requested page:

404.jsp:

<%@ page language="java" isErrorPage="true" contentType="text/html; charset=UTF-8" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
<head>
    <title>Error page</title>
    <meta charset="utf-8">
</head>
<body>
    <button onclick="history.back()">Back to Previous Page</button>
    <h1>404 Page Not Found.</h1>
    <br />
    <p><b>Error code:</b> ${pageContext.errorData.statusCode}</p>
    <p><b>Request URI:</b> ${pageContext.request.scheme}://${header.host}${pageContext.errorData.requestURI}</p>
    <br />
</body>
</html>

Useful reading:

"JSP.1.4 Error Handling" section in the JavaServer Pages Specification (JSR 245).
Handling JSP Page Errors (from the official Java EE 5 Tutorial)

P.S.
The use of scriptlets inside JSP is highly discouraged. Read this post.

this is perfect! I am using Spring MVC, just in 404,403,500 custom error pages I want to do some "dirty" work — , Jan 13 '13 at 04:02

How to show the requested URL in a JSP error page?

1 Answers1

Useful reading:

Linked