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When I am trying to start Apache Tomcat Server through cmd prompt using startup.bat, I am getting this error: 

JRE_HOME variable is not defined correctly. The environment variable is needed to Run this program

I have Defined the following Environment Variables Path as-

CATALINA_HOME - C:\Tomcat\apache-tomcat-7.0.78;

JAVA_HOME - C:\Program Files\Java\jdk1.8.0_65;

JRE_Home - C:\Program Files\Java\jre1.8.0_77;

CLASSPATH - C:\Program Files\Java\jdk1.8.0_65;\bin;C:\Program Files\Java\jre1.8.0_77\bin;C:\Tomcat\apache-tomcat-7.0.78\lib;
Mohd
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Amit
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  • I'm voting to close this question as off-topic because it is a duplicate and the original question is an easy find: https://stackoverflow.com/questions/7438160/java-home-variable – Marged May 31 '17 at 19:29

1 Answers1

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You can check the validity of the paths by

echo %JAVA_HOME%

echo %JRE_HOME%

echo %CATALINA_HOME%

Try to set CLASSPATH as

CLASSPATH=%JAVA_HOME%\bin;%JRE_HOME%\bin;%CATALINA_HOME%\lib

Community
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Dmitriy
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