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
Got the solution and it's working fine. Set the environment variables as:
CATALINA_HOME=C:\Program Files\Java\apache-tomcat-7.0.59\apache-tomcat-7.0.59
(path where your Apache Tomcat is)JAVA_HOME=C:\Program Files\Java\jdk1.8.0_25;
(path where your JDK is)JRE_Home=C:\Program Files\Java\jre1.8.0_25;
(path where your JRE is)CLASSPATH=%JAVA_HOME%\bin;%JRE_HOME%\bin;%CATALINA_HOME%\lib
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