Running servlet in tomcat

[ramdas pradeep kumar]

I am new to the Tomcat. I got a servlet project from someone else.
I've deployed the war file in the webapps dir.
It extracted the files and in the web-inf/classes dir and so many sub folders have been created as per the package structure. there is one servlet class in the innermost folder.
how can i call the servlet directly
any properties need to be changed in xml files, etc

[Malhar Barai]

Originally posted by ramdas pradeep kumar:
I am new to the Tomcat. I got a servlet project from someone else.
I've deployed the war file in the webapps dir.
It extracted the files and in the web-inf/classes dir and so many sub folders have been created as per the package structure. there is one servlet class in the innermost folder.
how can i call the servlet directly
any properties need to be changed in xml files, etc

Hi Pradeep
Do check the web.xml file, do you find any thing like startup tag, also check for the Servlet Context in server.xml file, you mite need to add a context..
hth
MB

[Faisal Khan]

I believe it can be done as following:
http://localhost:8080/myapp/servlet/com.abc.test.hello.MySerlvet
hth

[ramdas pradeep kumar]

hai malhar,
u mean the xml files in the config dir? can you explain me please.
i found the following in server.xml
<Context path="/examples" docBase="examples" debug="0"
reloadable="true" crossContext="true">

a segment from web.xml
<servlet>
<servlet-name>default</servlet-name>
<servlet-class>
org.apache.catalina.servlets.DefaultServlet
</servlet-class>
<init-param>
<param-name>debug</param-name>
<param-value>0</param-value>
</init-param>
<init-param>
<param-name>listings</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
[ September 19, 2003: Message edited by: ramdas pradeep kumar ]

[ritchie diputado]

Hi heres a sample xml configuration for servlet:
<servlet>
<servlet-name>myservlet</servlet-name>
<servlet-class>mypackage.anotherpackage.MyServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>myservlet</servlet-name>
<url-pattern>/myservlet</url-pattern>
</servlet-mapping>

Now you can call this serverlet by:
http://localhost:8080/mycontext/myservlet
Hope it helps
Ritchie Diputado

[Malhar Barai]

Originally posted by ramdas pradeep kumar:
hai malhar,
u mean the xml files in the config dir? can you explain me please.
i found the following in server.xml
<Context path="/examples" docBase="examples" debug="0"
reloadable="true" crossContext="true">

The context is in server.xml indeed in the "conf" directory...

a segment from web.xml
<servlet>
<servlet-name>default</servlet-name>
<servlet-class>
org.apache.catalina.servlets.DefaultServlet
</servlet-class>
<init-param>
<param-name>debug</param-name>
<param-value>0</param-value>
</init-param>
<init-param>
<param-name>listings</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>

The web.xml in the WEB-INF directory...
The following should be your directory structure
-bin
-conf
|- server.xml
|- tomcat-users.xml
|- web.xml(x - This is on the web.xml, we hv to use...)
-...
-webapps
|- examples(Contains examples)
|- <context you define>
|- WEB-INF(inside your context)
|- web.xml(inside WEB-INF)
-...
hth
MB
[ September 19, 2003: Message edited by: Malhar Barai ]
[ September 19, 2003: Message edited by: Malhar Barai ]