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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;
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Thanks in Advance. - JAVA_HOME: must point to installation directory of JDK. - JRE_HOME: must point to installation directory of JRE. - CLASSPATH: contains libraries path which JVM will look for.
set JAVA_HOME=C:\ "top level directory of your java install" set CATALINA_HOME=C:\ "top level directory of your Tomcat install"
In environment variables define CATALINA_HOME as a new variable and the value for the variable value assign to >>> C:\apache\apache-tomcat-7.0.40. in the path append this %CATALINA_HOME%\bin to get the command "catalina start" to work. Follow this answer to receive notifications.
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
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