TomCat HTTP Status 404 - /TestServlet

[Harry Tuttle]

I have a html-file named index.html with a send button that runs on Apache TomCat 7.0.2. When the send button is pushed a Servlet class named TestServlet should be executed, here call in the html-file

<form action="http://127.0.0.1:8080/TestServlet" method="POST" onSubmit='sign.value=signPad.getSign();'>   

The problem is, when then send button is pushed, TomCat can't find the Servlet TestServlet. The details about TestServlet are:

- the TestServlet is in a package named testclass
- the TestServlet is stored in ../Apache Software Foundation/Tomcat 7.0/webapps/Signpad/WEB-INF/classes/testclass/
- The web.xml is in the folder ../Apache Software Foundation/Tomcat 7.0/webapps/Signpad/WEB-INF and looks like

 <?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>MySignpad</display-name>
	<welcome-file-list>
		<welcome-file>index.html</welcome-file>
	</welcome-file-list>
	<servlet>
		<servlet-name>TestServlet</servlet-name>
		<servlet-class>testclass.TestServlet</servlet-class>
	</servlet>
	<servlet-mapping>
		<servlet-name>TestServlet</servlet-name>
		<url-pattern>/TestServlet</url-pattern>
	</servlet-mapping>
</web-app>   

i need help how to run TestServlet. can somebody help?

[Harry Tuttle]

Harry Tuttle wrote:
I have a html-file named index.html with a send button that runs on Apache TomCat 7.0.2. When the send button is pushed a Servlet class named TestServlet should be executed, here call in the html-file

<form action="http://127.0.0.1:8080/TestServlet" method="POST" onSubmit='sign.value=signPad.getSign();'>   

i could fix the problem by modifying the above form action in

<form action="http://127.0.0.1:8080/mywebapp/TestServlet" method="POST" onSubmit='sign.value=signPad.getSign();'>   

where mywebapp is the name of my web-application.

[Bear Bibeault]

Hard-coding the context path and using an absolute URL will get you ito trouble. Use a server-relative path (starting with the context path) and programmatically obtain the context path from the request.

See the JSP FAQ for details.