如何在startup.bat中设置多个JAVA_OPTS选项 [英] How to set multiple JAVA_OPTS options in startup.bat

问题描述

我尝试通过 startup.bat 启动tomcat时传递多个参数.我尝试将这些行添加到startup.bat文件的顶部，但是它们不起作用.

I am trying to pass multiple parameters when I start tomcat through startup.bat. I tried adding these lines at the top of startup.bat file, however they do not work.

set JAVA_OPTS="-Dapplication.home=E:\\webapp -Dfilepath=D:\\newFolder\\conf\\con.properties"

最初，我仅使用一个参数 -Dapplication.home = E:\\ webapp 运行该应用程序，但运行良好.现在，我需要传递另一个参数，此方法将失败.请指教.

Initially I was running the application with just one parameter -Dapplication.home=E:\\webapp which worked fine. Now I need to pass another parameter and this method fails. Please advice.

在运行时，我得到此异常 FileNotFoundException :

On running, I get this exception a FileNotFoundException:

java.io.FileNotFoundException: E:\webapp -Dfilepath=D:\newFolder\conf\con.properties (The filename, directory name, or volume label syntax is incorrect)
    at java.io.FileInputStream.open(Native Method)
    at java.io.FileInputStream.<init>(FileInputStream.java:120)
    at java.io.FileInputStream.<init>(FileInputStream.java:79)

代码将整个段作为单个参数读取.

The code is reading the entire segment as a single argument.

推荐答案

不带引号的尝试

set JAVA_OPTS=-Dapplication.home=E:\\webapp -Dfilepath=D:\\newFolder\\conf\\con.properties

应该工作

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