Hi, I got a test.java servlet in c:\temp subdirectory that needs to be compiled with -classpath option servlet.jar.
My classpath is as follows: SET CLASSPATH=.;%JAVA_HOME%\bin;%JAVA_HOME%\jre\bin My question is: What would be the exact command to compile the servet. I am running Windows 2000, with J2SE and TOMCAT Apache WebServer. Thanks in advance for any help.
To run a servlet the servlet.jar file should be in the classpath. my advise is while u r setting the classpath u r reffering to the bin direcotry there is the problem because bin directory is used to set the path not the classpath. In the classpath u have to set the .jar files etc..
for u r case u place the servlet.jar file in the c:\temp direcotry and run the follwoing command
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