Getting the dependecies classpath variable

[Paulo Silveira <pa...@paulo.com.br>]

Hello!

In my maven.xml, I have a goal that executes a class, which is inside
one of my dependecies´ jars.
How can I get this classpath? I ve tried with <pathelement
path="maven.dependency.classpath"/>, (or with
${}, but no success at all.

The snippet:

	<goal name="jmo:createdatabase">
		<java
classname="net.sf.hibernate.tool.hbm2ddl.SchemaExport" fork="yes">
			<arg
value="src/br/com/theacmeclan/Clan.hbm.xml"/>
			<arg
value="src/br/com/theacmeclan/User.hbm.xml"/>
			<arg value="--output=hb.sql"/>
			<arg
value="--properties=src/hibernate.properties"/>
			<classpath>
				<pathelement
path="maven.dependency.classpath"/>
			</classpath>
		</java>
	</goal>

thanks

---------------------
Paulo Silveira
http://www.paulo.com.br/
http://www.guj.com.br/
 

> -----Original Message-----
> From: Martin Skopp [mailto:skopp@riege.de] 
> Sent: quinta-feira, 8 de maio de 2003 12:46
> To: Maven Users
> Subject: Eclipse plugin: 2.1 specials + absolute source path
> 
> 
> Hi,
> 
> have a few questions/remarks to the eclipse plugin, the goal 
> to generate the .classpath file.  Two problems with the 
> output .classfile:
> 
> 1. path to sources (src/java) is absolute in the generated 
> .classfile. IMHO relative path would be better, especially 
> when you share .classfile via CVS (well, some people do in 
> some projects)
> 
> 2. Eclipse 2.1 (2.0 IMHO doesn't) allows to specify output 
> folder per source folder by a "output=" attribute.  If 2.0 
> ignores that tag, the plugin could use the tag to make 2.1 
> users more happy.
> 
> I get:
> ...
> <classpathentry kind="src" 
> path="/opt/cvsroot/myproject/bla/common/src/java">
> </classpathentry>
> <classpathentry kind="src" 
> path="/opt/cvsroot/myproject/bla/common/src/test">
> </classpathentry>
> ...
> 
> and my preferred output would be
> ...
> <classpathentry kind="src" path="src/java"/>
> <classpathentry kind="src" output="target/test-classes" 
> path="src/test"/> ...
> 
> 3. Just cosmetics, but isn't <classpathentry .../> nicer than 
> <classpathentry ...></classpathentry> ?
> 
> P.S.: Have seen the latest commit by jstrachan, but "output=" 
> would be even nicer since it makes eclipse to compile to the 
> right directory.
> 
> P.P.S: I can provide a path for 2. but I am too stupid to fix 
> 1. and 3.
> 
> Cheers,
> -- 
> Martin Skopp
> Riege Software International GmbH
> Support: mailto:maint@riege.com, Information: http://www.riege.com
> 
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