CODEWITHSUNDEEP
javaservlets
DarkKnight
DarkKnight

Reputation: 243

Error running web project in eclipse

I am a newbie to servlets. I created a dynamic web project in eclipse. I have following files in my project

home.html validateServlet.java

I have defined validated servlet as an action in home.html form. However when I run it, I get http status 404. Below is the hierarchy of my project

Project
Java Resources
   src
     com.servlets
    ValidateServlet.java
build
WebContent
    META-INF
    WEB-INF
        web.xml
    hello.html

Contents of my web.xml are as follows:

<?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>Website</display-name>
<servlet>
<description></description>
<display-name>ValidateServlet</display-name>
<servlet-name>validate</servlet-name>
<servlet-class>com.oracle.coen235.servlets.ValidateServlet</servlet-class>
</servlet>
</web-app>

In my hello.html, action is specified as ,

What might be the issue? I guess I am not able to generate the class file for my servlet. Can anyone guide me through this problem?

Upvotes: 0

Views: 57

Answers (1)

Udo Held
Udo Held

Reputation: 12538

You seem to mix things up.

<servlet-class>com.oracle.coen235.servlets.Validate</servlet-class>

Should list the full class name. I don't see your package name.

It should be

<servlet-class>com.oracle.coen235.servlets.ValidateServlet</servlet-class>

or whatever your package name is.

We are you mapping your servlet to a path? It should be something like:

 <servlet-mapping>
      <servlet-name>validate</servlet-name>
      <url-pattern>/validate</url-pattern>
 </servlet-mapping>

or an annotation directly in your servlet class.

Upvotes: 1

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