How to set CATALINA_HOME variable in windows 7?

I have downloaded apache-tomcat-7.0.35. My JDK version is jdk1.6.0_27.

How do I configure CATALINA_HOME as an environment variable and how do I run Tomcat server under Windows 7?

Setting the JAVA_HOME, CATALINA_HOME Environment Variable on Windows

One can do using command prompt:

1. set JAVA_HOME=C:\ "top level directory of your java install"
2. set CATALINA_HOME=C:\ "top level directory of your Tomcat install"
3. set PATH=%PATH%;%JAVA_HOME%\bin;%CATALINA_HOME%\bin

OR you can do the same:

1. Go to system properties
2. Go to environment variables and add a new variable with the name JAVA_HOME and provide variable value as C:\ "top level directory of your java install"
3. Go to environment variables and add a new variable with the name CATALINA_HOME and provide variable value as C:\ "top level directory of your Tomcat install"
4. In path variable add a new variable value as ;%CATALINA_HOME%\bin;

Assuming Java (JDK + JRE) is installed in your system, do the following steps:

1. Install Tomcat7
2. Copy 'tools.jar' from 'C:\Program Files (x86)\Java\jdk1.6.0_27\lib' and paste it under 'C:\Program Files (x86)\Apache Software Foundation\Tomcat 7.0\lib'.
3. Setup paths in your Environment Variables as shown below:

C:>echo %path%

C:\Program Files (x86)\Java\jdk1.6.0_27\bin;%CATALINA_HOME%\bin;

C:>echo %classpath%

C:\Program Files (x86)\Java\jdk1.6.0_27\lib\tools.jar;
C:\Program Files (x86)\Apache Software Foundation\Tomcat 7.0\lib\servlet-api.jar;

C:>echo %CATALINA_HOME%

C:\Program Files (x86)\Apache Software Foundation\Tomcat 7.0;

C:>echo %JAVA_HOME%

C:\Program Files (x86)\Java\jdk1.6.0_27;

Now you can test whether Tomcat is setup correctly, by typing the following commands in your command prompt:

C:/>javap javax.servlet.ServletException
C:/>javap javax.servlet.http.HttpServletRequest

It should show a bunch of classes

Now start Tomcat service by double clicking on 'Tomcat7.exe' under 'C:\Program Files (x86)\Apache Software Foundation\Tomcat 7.0\bin'.