CATALINA_HOME 环境变量定义不正确
我将 Apache Tomcat 6.0 安装到 C:/Program Files 文件夹，并设置了环境变量 JAVA_HOME 和 CATALINA_HOME，但是当我启动服务器时使用 startup.bat，我收到错误：
CATALINA_HOME 环境变量定义不正确。
我的 CATALINA_HOME 环境变量设置为 CATALINA_HOME=C:\Program Files\apache-tomcat-6.0.35。
我做错了什么？
I installed Apache Tomcat 6.0 to a C:/Program Files folder and I set the environmental variables JAVA_HOME and CATALINA_HOME, but when I start the server using startup.bat, I am getting the error:
CATALINA_HOME environmental variable is not defined correctly.
My CATALINA_HOME environment variable is set to CATALINA_HOME=C:\Program Files\apache-tomcat-6.0.35.
What did I do wrong?
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首先你必须提到tomcat，直到tomcat的版本参考下面
CATALINA_HOME: C:\Program Files\apache-tomcat-6.0.35;
然后将 CLASSPATH 变量设置为
CLASSPATH : C:\Program Files\apache-tomcat-6.0.35\lib;
然后最后将路径变量编辑为
路径：.;%CATALINA_HOME%\bin;
注意：必须在每个变量的末尾添加分号(;)。
At first you have to mention tomcat up to tomcat's version refer below
CATALINA_HOME: C:\Program Files\apache-tomcat-6.0.35;
then set CLASSPATH variable to
CLASSPATH : C:\Program Files\apache-tomcat-6.0.35\lib;
then finally edit path variable to
PATH: .;%CATALINA_HOME%\bin;
NOTE: a semicolon(;) must be put at the end of each and every variable.
在调用 service.bat 文件之前，请确保将 cmd 窗口上的目录 cd 到 Tomcat 所在的路径
Make sure you cd your directory on the cmd window to the path where the Tomcat is before you you call the service.bat file
打开 catalina.bat / catalina.sh .. 并检查 CATALINA_HOME ，如果您从某处复制了 Tomcat，则这可能已被硬编码。
Open up your catalina.bat / catalina.sh .. and check for CATALINA_HOME , this could have been hardcoded if you have copied Tomcat from somewhere.
的路径位置
我也遇到了同样的问题，我复制了 JRE_HOME JRE_Home= "C:Progra******** ;"
在这里我输入了分号，我们不应该输入分号，我删除了它，
然后Tomcat启动成功。
所以设置路径后必须去掉末尾的分号
JRE_HOME 或
JAVA_HOME 或
CATALINA_HOME 。
I have got the same problem, I copied the path location of JRE_HOME
JRE_Home= "C:Progra******** ;"
Here i have entered Semicolon which we should not , I removed it,
then Tomcat Started successfully .
So Semicolon must be removed at the end after setting path
JRE_HOME or
JAVA_HOME or
CATALINA_HOME .
您还需要将 JAVA_HOME 变量设置为指向 JDK，而不是 JRE。
Tomcat 必须知道 Java 是否已安装。它需要 JDK。
最后重启电脑。
https://serverfault.com/questions/ 579159/the-catalina-home-environment-variable-is-未正确定义-this-environment
You also need to set the JAVA_HOME variable to point to the JDK, not JRE.
Tomcat must know were Java is installed. It needs the JDK.
In the end restart computer.
https://serverfault.com/questions/579159/the-catalina-home-environment-variable-is-not-defined-correctly-this-environment
请注意删除定义用户变量末尾的分号，例如 JAVA_HOME、CATALINA_HOME
Please care to remove the semi colon at the end of defining user variables such as JAVA_HOME, CATALINA_HOME
在环境变量中将 CATALINA_HOME 定义为一个新变量，并将该变量的值分配给 >> C:\apache\apache-tomcat-7.0.40
在路径中附加此 %CATALINA_HOME%\bin 以使命令“catalina start”起作用
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
设置
CATALINA_HOME：C:\apache-tomcat-9.0.27
Tomcat目录的路径（无需添加“bin”）
CLASSPATH : C:\apache-tomcat-9.0.27\lib
对于 Windows 10 较新版本，删除“;”并更新 PATH 变量的每个条目
Set the
CATALINA_HOME : C:\apache-tomcat-9.0.27
The path of the Tomcat directory (no need to add "bin")
CLASSPATH : C:\apache-tomcat-9.0.27\lib
For Windows 10 newer versions, remove the ";" and update each entry to the PATH variable