parameter passing from form to servlet

[Dhanashree Mankar]

my servlet file is in
C:\tomcat4\webapps\examples\WEB-INF\classes\coreservlets\parameters.java
and html file is in
C:\tomcat4\webapps\forms\parametersform.html
Then i tried to access this form as follows
127.0.0.1:8080/forms/parametersform.html
it gives 404 error
Html file has action as follows:
<form action="/servlet/coreservlets.parameters">
plz help

[shankar vembu]

Originally posted by Dhanashree Mankar:
my servlet file is in
C:\tomcat4\webapps\examples\WEB-INF\classes\coreservlets\parameters.java

So your base dir is C:\tomcat4\webapps\examples\
And you need to have all your html files in here.
Copy your forms folder to C:\tomcat4\webapps\examples\
and then try http://127.0.0.1:8080/forms/parametersform.html
should work.
regards.

[Sim Raina]

try this
127.0.0.1:8080/parametersform.html
and when calling servlet
127.0.0.1:8080/servlet/servletname

[William Brogden]

my servlet file is in
C:\tomcat4\webapps\examples\WEB-INF\classes\coreservlets\parameters.java
and html file is in
C:\tomcat4\webapps\forms\parametersform.html

1. parameters.java - you need to compile it to parameters.class
2. \forms\parameterform.html - If you do not have a forms\WEB-INF directory and a web.xml in that directory, Tomcat will NOT recognize forms as a valid web application and will not serve the html page.
3. use of /servlet/ in the action - IEEEEEE the invoker servlet problem strikes again. Read about it here in the excellent FAQ.
Bill