Show only existing executable dirs in PATH using only builtin bash commands
~/askapache/bin ~/askapache/sbin /bin /etc /sbin /usr/X11R6/bin /usr/bin /usr/bin/mh /usr/libexec /usr/local/bin /usr/sbin /etc/X11 /etc/X11/xinit ~/askapache/ruby/bin /usr/local/dh/apache/template/bin /usr/local/dh/apache2/template/bin /usr/local/dh/apache2/template/build /usr/local/dh/apache2/template/sbin /usr/local/dh/bin /usr/local/dh/java/bin /usr/local/dh/java/jre/bin /usr/local/php5/bin .
What do you think?
Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?
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GP=; for p in ${PATH//:/ }; do [[ -d $p && -x $p ]] && [[ "${GP}s" == s ]] && GP=$p || GP=${GP}:$p; done; export PATH=$GP