When trying to start Tomcat Server through cmd prompt using 'startup.bat' getting error as-"JRE_HOME variable is not defined correctly. The environment variable is needed to Run this program" Defined Environment path as-
CATALINA_HOME-C:\Program Files\Java\apache-tomcat-7.0.59\apache-tomcat-7.0.59
JAVA_HOME-C:\Program Files\Java\jdk1.8.0_25;
JRE_Home-C:\Program Files\Java\jre1.8.0_25\bin;
This question is related to
java
apache
tomcat
batch-file
Your JRE_HOME does not need to point to the "bin" directory. Just set it to C:\Program Files\Java\jre1.8.0_25
Source: Stackoverflow.com