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Replies:
6
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Last Post:
Jun 3, 2007 10:55 AM
by: greeneyed
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Launching a java application under windows
Posted:
May 31, 2007 2:53 PM
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Our product uses a batch file to launch our swing app. We are running into a problem though because Windows 2000 has a command line limit of 2047 and our classpath is starting to hit this limit. Is there some other way to pass the classpath to the JVM?
Thanks,
Aberrant
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Re: Launching a java application under windows
Posted:
May 31, 2007 7:59 PM
in response to: aberrant
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Correct |
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Put the classpath into the Class-Path attribute of the manifest of your main jar file. Then your batch file can be as simple as this:
java -jar main.jar
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Re: Launching a java application under windows
Posted:
Jun 1, 2007 3:31 AM
in response to: twalljava
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I think that will work, but our installer places files in different places, depending on the platform. I think I'm going to encourage the install person to keep them all in the same relative location. Thanks.
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Re: Launching a java application under windows
Posted:
Jun 1, 2007 5:30 AM
in response to: aberrant
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Helpful |
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Your manifest entry can include directories too.. but only jars will be loaded when you just specific a directory.
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Re: Launching a java application under windows
Posted:
Jun 1, 2007 6:17 AM
in response to: aberrant
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Or you can use the JarJar Ant task to unpack all your jars and repack them into a single jar that contains your entire application.
Example from JRuby's build.xml:
<target name="jar-complete" depends="generate-method-classes" description="Create the 'complete' JRuby jar. Pass 'mainclass' and 'filename' to adjust.">
<property name="mainclass" value="org.jruby.Main"/>
<property name="filename" value="jruby-complete.jar"/>
<taskdef name="jarjar" classname="com.tonicsystems.jarjar.JarJarTask" classpath="${lib.dir}/jarjar-0.7.jar"/>
<jarjar destfile="${lib.dir}/${filename}">
<fileset dir="${jruby.classes.dir}">
<exclude name="org/jruby/util/ant/**/*.class"/>
</fileset>
<fileset dir="${basedir}/lib/ruby/1.8">
<include name="**/*.rb"/>
</fileset>
<zipfileset src="${lib.dir}/asm-2.2.3.jar"/>
<zipfileset src="${lib.dir}/asm-commons-2.2.3.jar"/>
<zipfileset src="${lib.dir}/jline-0.9.91.jar"/>
<zipfileset src="${lib.dir}/backport-util-concurrent.jar"/>
<rule pattern="org.objectweb.asm.**" result="jruby.objectweb.asm.@1"/>
<zipfileset dir="${basedir}" prefix="META-INF/jruby.home">
<include name="bin/*"/>
<include name="lib/ruby/gems/1.8/cache/sources*.gem"/>
<include name="lib/ruby/gems/1.8/gems/sources*/**/*"/>
<include name="lib/ruby/gems/1.8/specifications/sources*.gemspec"/>
<include name="lib/ruby/site_ruby/**/*"/>
</zipfileset>
<manifest>
<attribute name="Built-By" value="${user.name}"/>
<attribute name="Main-Class" value="${mainclass}"/>
</manifest>
</jarjar>
</target>
JarJar lives on SourceForge.net: http://sourceforge.net/projects/jarjar
Message was edited by: fullung
Message was edited by: fullung
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Re: Launching a java application under windows
Posted:
Jun 1, 2007 7:19 PM
in response to: aberrant
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Forget about batch file. Try this :-
jsmooth.sf.net
it give you .exe file
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Re: Launching a java application under windows
Posted:
Jun 3, 2007 10:55 AM
in response to: aberrant
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Helpful |
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If/when you are lucky enough to be able to use Java 6, you will be able to use * in directories with .jar files to include them all.
That should help.
OTOH, until that, the Class-Path seems to be the way to go. Just be aware that once you use java -jar whatever.jar, "-cp" and the CLASSPATH environment do nothing and everything has to be specified through the manifest. Just in case  .
S!
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