title: how to start apache-tomcat- 5.0.28 in windows 7 to run servlet Program in java
body: <p>I have installed downloaded and extracted apache-tomcat- 5.0.28 in windows 7 Operating system. and then I followed below given instruction.</p>
  
  <ol>
  <li>Now Right click on my computer icon, Select properties->Advanced->Environment Variables. </li>
  <li>Click on new Button in System Variable frame (highlighted in above figure). Clicking on new button open dialog box to add new System variables. </li>
  <li>Now type CATALINA_HOME in variable name and C:\Program
  Files\jakarta-tomcat-5.0.28 in variable value and then click ok.</li>
  <li>Again click on new Button in System Variable frame</li>
  <li>Now type JAVA_HOME in variable name and C:\Program Files\Java\jdk1.5.0_21 in variable value and then click ok.</li>
  <li>Now select path variable from System Variable frame
  click on edit Button in System Variable frame 
  now add ; C:\Program Files\apache-tomcat-5.0.28\lib; and then click ok.</li>
  <li>Now open Dos prompt by clicking start->run->cmd
  type SET CLASSPATH= C:\Program Files\jakarta-tomcat-5.0.28\lib\servlet-api.jar
  type cd C:\Program Files\jakarta-tomcat-5.0.28\bin
  type startup (to start tomcat server) and when you need to shutdown server type shutdown</li>
  </ol>
  
  <p>After completing all above step when I type startup in dos prompt I got tons of errors and could not start apache tomcat.
  Can any one suggest any solution?</p>