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Full Discussion: escaping backslashes to evaluate paths
Top Forums Shell Programming and Scripting escaping backslashes to evaluate paths Post 302419944 by panyam on Monday 10th of May 2010 07:11:26 AM
Old 05-10-2010
Something like this:

Code:
 
echo "CLASSPATH=/opt/app/example.jar" | sed 's!/opt/app/example.jar!/opt/app/default.jar!'

 

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LEARN ABOUT SUSE

build-jar-repository

BUILD-JAR-REPOSITORY(1) 					   User Commands					   BUILD-JAR-REPOSITORY(1)

NAME

       build-jar-repository - create a symbolic link to a JAR

SYNOPSIS

       build-jar-repository [OPTION]... DIRECTORY... JAR

DESCRIPTION

       Build a JAR repository in the named directory by copying files or creating symbolic links

OPTIONS

       If no option is specified the default action will be to create symbolic links

       -c, --copy
	      copy files

       -h, --hard
	      create hard links

       -p, --preserve-naming
	      try  to  preserve  the  names  of  the original JAR files (in case of a nested hit the slashes in the path will still be replaced by
	      underscores)

	      using this option makes any future automated repository rebuild impossible, and implies -c unless specified otherwise

       -s, --soft, --symbolic
	      create symbolic links (default)

       --help
	      display help text

EXAMPLES

       build-jar-repository . jndi
	      will create a symbolic link to the JNDI JAR in the current working directory

       build-jar-repository -h /tmp oro
	      will create a hard link to the ORO JAR in /tmp

AUTHOR

       Written by Nicholas Mailhot and David Walluck

REPORTING BUGS

       Report bugs using JPackage Bugzilla (http://www.jpackage.org/bugzilla/)

build-jar-repository (jpackage-utils) 1.7.0			    March 2006						   BUILD-JAR-REPOSITORY(1)
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