Compiling a servlet in command line

Hi,
Can someone show me the command to type in to compile a servlet in command line. I know i have to include the servlet.jar file, but i don't remember the exact command.
Thanks

D:\logs>javac
Usage: javac <options> <source files>
where possible options include:
-g Generate all debugging info
-g:none Generate no debugging info
-g:{lines,vars,source} Generate only some debugging info
-nowarn Generate no warnings
-verbose Output messages about what the compiler is doing
-deprecation Output source locations where deprecated APIs are used
-classpath <path> Specify where to find user class files
-sourcepath <path> Specify where to find input source files
-bootclasspath <path> Override location of bootstrap class files
-extdirs <dirs> Override location of installed extensions
-d <directory> Specify where to place generated class files
-encoding <encoding> Specify character encoding used by source files
-source <release> Provide source compatibility with specified release
-target <release> Generate class files for specific VM version
-help Print a synopsis of standard options

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    Ben

    As long as you are not more specific about the steps you take to verify that the ears are the same, I cannot offer any more help. All I can say is: your assumption is wrong, because if they were the same, it would work. Simple logic is usually the best.

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    Thanks for the answer, Carlos.
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