Hi all, I use Tomcat 3.2.3 for test servlet.
The default path for tomcat (http://localhost:8080) is /home/tomcat/webapps/ROOT/index.html Of COURSE, http://localhost:8080/servlet/SnoopServlet is work. However I put a simple html file in the same directory (classes) as follows == <FORM ACTION="./SnoopServlet" METHOD="post"> �R�O�G <SELECT NAME=SCommand> <OPTION SELECTED>testyou <OPTION>test1 <OPTION>test2 <OPTION>test3 </SELECT> == It does not work. What I miss?!?!?? -->Anyway, -->I just want to write a html file and be ACTION to a servlet. -->what is the path SHOULD I put the files (.html and servler program) I am confused of the default path of SERVLET program. thanks all! --Sam ___________________________________________________________________________ To unsubscribe, send email to [EMAIL PROTECTED] and include in the body of the message "signoff SERVLET-INTEREST". Archives: http://archives.java.sun.com/archives/servlet-interest.html Resources: http://java.sun.com/products/servlet/external-resources.html LISTSERV Help: http://www.lsoft.com/manuals/user/user.html
