从 windows 命令行运行 Java

Question

I recently asked a bout a java program I could not find the start point to and was told to look in the bat file which started it but I do not understand what it says.

This program runs on a tomcat server and must be running before the.jsp pages will run so they are not the starting point.

Can anyone explain what this means and where the program starts so I can do a step through.

D:\foo_development\server>java -Xmx256M -Djava.library.path=lib -Djava.rmi.serv
er.hostname=192.168.0.104 -Djava.rmi.server.codebase="file:/D:\foo_development\
server\foo_server.jar http://192.168.0.104:8000/foo_client_stubs/server_keycon
sole_stubs.jar" -Djava.security.policy=conf/java.policy -jar foo_server.jar 109
8 2001

This is exactly how it appears in the cmd window after I click run.bat

Answer 1

Let's analyze the command line:

• java : the executable
• -Xmx256M : how much memory to use
• -Djava.library.path=lib where to find native libraries (for JNI)
• -Djava.rmi.server.hostname=192.168.0.104 -Djava.rmi.server.codebase="file:/D:\foo_development\ server\foo_server.jar http://192.168.0.104:8000/foo _client_stubs/server_keycon sole_stubs.jar" RMI related stuff
• -Djava.security.policy=conf/java.policy which security policy to use
• -jar foo_server.jar the JAR file to execute! This is the juicy bit!
• 109 8 2001 arguments passed to your main() method

So foo_server.jar is the.jar file that gets executed. Check its META-INF/MANIFEST.MF for a line that starts with Main-Class: and that should tell you which class gets executed.

Answer 2

D:\foo_development\server>java -Xmx256M -Djava.library.path=lib -Djava.rmi.serv
er.hostname=192.168.0.104 -Djava.rmi.server.codebase="file:/D:\foo_development\
server\foo_server.jar http://192.168.0.104:8000/foo_client_stubs/server_keycon
sole_stubs.jar" -Djava.security.policy=conf/java.policy -jar foo_server.jar 109
8 2001

This command runs foo_server.jar file with arguments 109,8,201. And all other before -jar are vm arguments to tell jvm about the different memory size like heapmemory size and aslo library path.

For your main class you will find its entry in MANIFIST file as Main-class:-- foo.bar.MainClass

Answer 3

In short you have

java -jar foo_server.jar 

The rest is just configuration and arguments.

In the foo_server.jar there will be a MANIFEST.MF file which contains a line like

Main-Class: com.mycompany.foo.FooMain

This is the class where main(String[]) is called when it is loaded.

Answer 4

Looking at the command:

• java is the executable to run ( java.exe , found on the path)
• -Xmx256M specifies a maximum heap size of 256 megabytes
• -D (multiple times) defines various Java system properties
• -jar means "load the given jar file, and start with the class specified by the Main-Class attribute in its manifest
• foo_server.jar is the jar file to load
• 1098 2001 are the command line arguments to main (so args will be an array with elements "1098" and "2001"

Answer 5

The key part is the options -jar foo_server.jar . This tells the JRE to open that named JAR, to read the Main-Class property from its metadata (in the archive's META-INF/MANIFEST.MF ) and execute that (starting from its static main method, as normal).

Answer 6

The main class of an jar can be specified in the manifest of this jar. (@see http://download.oracle.com/javase/tutorial/deployment/jar/appman.html )

In your case it is the foo_server.jar . This jar (a jar it is just a zip) must have a folder META-INF where the manifest is located.