Hi Everyone,

I believe I do not understand how to configure a servlet in the webapp's web.xml hence need a hand conceptually. I get the following error when I added an additional servlet to my webapp:


org.xml.sax.SAXParseException: The content of element type "web-app" must match "(icon?,display-name?,description?,distributable?,context- param*,filter*,filter-mapping*,listener*,servlet*,servlet- mapping*,session-config?,mime-mapping*,welcome-file-list?,error- page*,taglib*,resource-env-ref*,resource-ref*,security- constraint*,login-config?,security-role*,env-entry*,ejb-ref*,ejb-local- ref*)".
at org.apache.xerces.util.ErrorHandlerWrapper.createSAXParseException(Error HandlerWrapper.java:232)


Here's a snipplet of the webapp's web.xml:

<?xml version="1.0" encoding="ISO-8859-1"?>

<!DOCTYPE web-app
PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd";>

<web-app>

<display-name>Tomcat Examples</display-name>
<description>
Tomcat Example servlets and JSP pages.
</description>

<servlet>
<servlet-name>C2</servlet-name>
<servlet-class>C2</servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>C2</servlet-name>
<url-pattern>/controller</url-pattern>
</servlet-mapping>

<servlet>
<servlet-name>C3</servlet-name>
<servlet-class>C3</servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>C3</servlet-name>
<url-pattern>/helper</url-pattern>
</servlet-mapping>





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