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path manipulation in bash

Terminal - path manipulation in bash
rp() { local p; eval p=":\$$1:"; export $1=${p//:$2:/:}; }; ap() { rp "$1" "$2"; eval export $1=\$$1$2; }; pp() { rp "$1" "$2"; eval export $1=$2:\$$1; }
2010-07-15 18:52:01
User: cout
Functions: eval export
0
path manipulation in bash

I used to do a lot of path manipulation to set up my development environment (PATH, LD_LIBRARY_PATH, etc), and one part of my environment wasn't always aware of what the rest of the environment needed in the path. Thus resetting the entire PATH variable wasn't an option; modifying it made sense.

The original version of the functions used sed, which turned out to be really slow when called many times from my bashrc, and it could take up to 10 seconds to login. Switching to parameter substitution sped things up significantly.

The commands here don't clean up the path when they are done (so e.g. the path gets cluttered with colons). But the code is easy to read for a one-liner.

The full function looks like this:

remove_path() { eval PATHVAL=":\$$1:" PATHVAL=${PATHVAL//:$2:/:} # remove $2 from $PATHVAL PATHVAL=${PATHVAL//::/:} # remove any double colons left over PATHVAL=${PATHVAL#:} # remove colons from the beginning of $PATHVAL PATHVAL=${PATHVAL%:} # remove colons from the end of $PATHVAL export $1="$PATHVAL" } append_path() { remove_path "$1" "$2" eval PATHVAL="\$$1" export $1="${PATHVAL}:$2" } prepend_path() { remove_path "$1" "$2" eval PATHVAL="\$$1" export $1="$2:${PATHVAL}" }

I tried using regexes to make this into a cleaner one-liner, but remove_path ended up being cryptic and not working as well:

rp() { eval "[[ ::\$$1:: =~ ^:+($2:)?((.*):$2:)?(.*):+$ ]]"; export $1=${BASH_REMATCH[3]}:${BASH_REMATCH[4]}; };

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Terminal - Alternatives

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What others think

Any idea how to format the full function so it looks right?

Comment by cout 214 weeks and 2 days ago

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