Can't Compile

Sorry for this totally noob question.
I can't compile a servlet.
What do I have to do to be able to compile a servlet?
What do I have to add to my classpath?

How are you trying to compile the code? What error messages do you get?
Copy and paste them here.

I am very sorry. But I was getting errors like can't find symbol and javax.servlet.* not found, etc.

But when I tried to re-open command prompt and compile again, I am suddenly able to.

I shall include my class-path here for everyone's benefit, and mark this thread as solved.

Thanks to Norm who replied.

But it would still be nice to know what is needed to compile a servlet. I have attached an image of the java stuff I have in Program Files.

I also include below my classpath under Windows XP->My Computer->Environment Variables:
C:\Program Files\PC Connectivity Solution\;%SystemRoot%\system32;%SystemRoot%;%Syst emRoot%\System32\Wbem;C:\Program Files\Java\jdk1.6.0_07\bin;C:\Program Files\Microsoft SQL Server\90\Tools\binn\;C:\Program Files\Microsoft SQL Server\80\Tools\Binn\;C:\0 Applications\jEdit;C:\Program Files\Java\jswdk-1.0.1\lib

Thanking all,
Siddharth

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what is needed to compile a servlet.

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There are classes defined in some jar file that are necessary to compile them.

Which JAR file Norm? The one in JWSDK or JSDK?

Quote:

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can't find symbol and javax.servlet.* not found

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The one(s) that contains the above.

With what software do I see what's inside a JAR file?

A jar file is almost the same as a zip file. Any utility that can read a zip should be able to read a jar file.