在Windows 7上的Tomcat中安装

我正在寻找安装Tomcat的方法. 因此，我解压缩了文件，将CATALINA_HOME设置为在命令行中看到的路径名(计算机上的c:\Users\myName\apache-tomcat-7.0.40) 我的 setenv.bat 如下所示:

I am looking to install Tomcat. So, I unzipped the file, set CATALINA_HOME to the path name I see on the command line (c:\Users\myName\apache-tomcat-7.0.40 on my machine) and my setenv.bat looks like as follows:

set "JRE_HOME=C:\Program Files\Java\jre6\bin;"
set "JAVA_HOME=C:\Program Files\Java\jdk1.7.0_07\bin;"
exit /b 0

我也尝试过

set "JRE_HOME=C:\Program Files\Java\jre6;"
set "JAVA_HOME=C:\Program Files\Java\jdk1.7.0_07;"
exit /b 0

还有

set "JRE_HOME=C:\Program Files\Java\jre6;"
exit /b 0

但是，当我调用 startup.bat 时，我不断收到错误消息"JRE_HOME环境变量未正确定义...".

However, when I invoke startup.bat, I keep getting the error that "The JRE_HOME environment variable is not defined properly...".

那里的路径名C:\Program Files\Java\jre6是Windows/系统环境变量的PATH变量上的路径名. JRE和JDK在这台计算机上运行了几个月以来，它们都安装得很好.

the path name there C:\Program Files\Java\jre6 is the one on my PATH variable of Windows/System Environment variables. JRE and JDK have been running since months on this machine-- they are installed fine.

我在哪里错了?

注意:我看到安装了Tomcat 7.0.37 ;在许多其他Tomcat安装问题中，未正确定义CATALINA_HOME .