Java 获取错误:尝试运行 Apache-Tomcat 的 startup.bat 时未正确定义 JRE_HOME 变量
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Getting Error:JRE_HOME variable is not defined correctly when trying to run startup.bat of Apache-Tomcat
提问by Rajeev Dixit
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-
当尝试使用 'startup.bat' 通过 cmd 提示符启动 Tomcat 服务器时,出现错误为-“JRE_HOME 变量未正确定义。运行此程序需要环境变量”定义的环境路径为-
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;
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;
采纳答案by Rajeev Dixit
Got the solution and it's working fine. Set the environment variables as:
得到了解决方案,它工作正常。将环境变量设置为:
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
CATALINA_HOME=C:\Program Files\Java\apache-tomcat-7.0.59\apache-tomcat-7.0.59
(您的 Apache Tomcat 所在的路径)JAVA_HOME=C:\Program Files\Java\jdk1.8.0_25;
(你的JDK所在的路径)JRE_Home=C:\Program Files\Java\jre1.8.0_25;
(您的 JRE 所在的路径)CLASSPATH=%JAVA_HOME%\bin;%JRE_HOME%\bin;%CATALINA_HOME%\lib
回答by Hypino
Your JRE_HOME does not need to point to the "bin" directory. Just set it to C:\Program Files\Java\jre1.8.0_25
您的 JRE_HOME 不需要指向“bin”目录。只需将其设置为C:\Program Files\Java\jre1.8.0_25