CODEWITHSUNDEEP
javajspservlets
petfreshman
petfreshman

Reputation: 523

My jsp cannot find my servlet, giving me a 404

I am trying to submit a simple form for processing in a servlet, but I keep getting a 404 whenever I submit. The code compiles and deploys fine. I've been through just about every tutorial on the subject, so I understand that there is a lot of information out there about it. Any advice on which rookie mistake I am making? Thanks in advance for any help!

jsp is located in /webapps/registerWidget servlet located in /webapps/registerWidget/WEB-INF/classes/com/sample/applet web.xml located in /webapps/registerWidget/WEB-INF

Here is my jsp:

    <!DOCTYPE HTML>
    <html>
    <head>
        <title>Sample Applet</title>
    </head>
    <body>
        <header>Please Register</header>
        <section>
            <form action="/register" method="POST">
                First name: <input type="text" name="firstname"><br>
                Last name: <input type="text" name="lastname"><br>
                Email: <input type="text" name="email"><br>
                <input type="submit" value="Submit">
            </form>
        </section>
    </body>
</html>

Here is my servlet:

package com.sample.applet;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class RegisterServlet extends HttpServlet {

    public void doGet(HttpServletRequest req, HttpServletResponse res)
            throws IOException, ServletException {
        doPost(req, res);
    }

    public void doPost(HttpServletRequest req, HttpServletResponse res)
            throws ServletException {

        String first = req.getParameter("firstname");
        String last = req.getParameter("lastname");
        String email = req.getParameter("email");

        ....some processing code....

    }
}

Here is my web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
    http://java.sun.com/xml/ns/j2ee/Web-app_2_4.xsd">

    <servlet>
        <servlet-name>registerServlet</servlet-name>
        <servlet-class>com.sample.applet.RegisterServlet</servlet-class>
    </servlet>

    <servlet-mapping>
        <servlet-name>registerServlet</servlet-name>
        <url-pattern>/register</url-pattern>
    </servlet-mapping>
</web-app>

Upvotes: 2

Views: 10499

Answers (2)

VishalDevgire
VishalDevgire

Reputation: 4268

Use 'register' in action instead of '/register'

form action="register"

'/' in '/register' tells it to search your servlet at root of your context.

Upvotes: 0

Vincent Ramdhanie
Vincent Ramdhanie

Reputation: 103135

The url in the action in your form starts with a slash. That implies that you are starting from the server root and looking for a servlet named register at that point. However, your servlet is actually in a context of some sort. So you can either say: /mycontext/register or remove the slash altogether.

Upvotes: 2

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