How to compile a servlet?

Question

tomcat

I want to compile a java file through tomcat, it doesn't work?

Environment variables set: JAVA_HOME: C:\Program Files\Java\jdk1.8.0_20
PATH: C:\Program Files\Java\jdk1.8.0_20\bin
Javac command works.

Answer 1

Remove the % sign before javac, and there will be happiness (if, of course, the other paths are correct).
If it is unclear why: the example is given for a unix-like system with some csh or tcsh shell. This can already be seen at least from the file delimiters (forward slash). And the % sign simply symbolizes that the javac command is being entered in response to a shell prompt, which it is. Under Windows, of course, you don’t need to write it :)

Answer 2

1) Compile from under the environment.
2) If you want from the CS, then the javac command does not work for you. You even wrote about it in Russian.
3) In Tomcat, you only run the servlet, you don't compile it. Here you are using the Kota library to compile your class.

Answer 3

Still not working, what should I do?