Modified PATH without /usr/bin

Question

How can I save a modified PATH in PATH_MOD that does not contain /usr/bin?

Output of PATH:

/opt/conda/bin:/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin

MakeReal, please show us in your question how you "tried it with sed". It might be that you're almost there and fixing your attempt would be better for your learning than providing a completely different solution — Chris Davies, May 27 '20 at 11:26

score 0 · Answer 1 · answered May 29 '20 at 17:47

0

have you tried:
PATH_MOD=$(echo $PATH | sed 's/:\/usr\/bin:/:/g')

EDIT: double \ in order to make backslashes visible

answered May 29 '20 at 17:47

Francesco Orlando

26

1

That wouldn't work if /usr/bin was first or last in $PATH or if there were 2 consecutive /usr/bin (as in PATH=/x:/usr/bin:/usr/bin:/y) – Stéphane Chazelas May 29 '20 at 18:21
1

That would also potentially not work properly if $PATH contained $IFS characters or wildcards or backslashes or was something like -Ennen... – Stéphane Chazelas May 29 '20 at 18:35

Stéphane Chazelas · Answer 2 · 2020-05-29T18:34:17.133

In the zsh shell:

path=("${path[@]:#/usr/bin}")

updates $PATH in place. Or to set $PATH_MOD instead:

PATH_MOD=${(j[:])path:#/usr/bin}

In zsh, $PATH is tied to the $path array (like in (t)csh) and ${array:#pattern} expands to the elements of the array that don't match the pattern.

Beware that if $PATH was just /usr/bin, then it becomes empty. For zsh, that means there's no command to be found anywhere, but for most of everything else, that means the current working directory is looked for for commands!

The equivalent in bash 4.4+ could be done using some helper function like:

remove_element() {
  local - text=$1 IFS=$2 element=$3 i
  set -o noglob
  set -- $text""
  for i do
    if [ "$i" != "$element" ]; then
       set -- "$@" "$i"
    fi
    shift
  done
  REPLY="$*"
}

remove_element "$PATH" : /usr/bin
PATH_MOD=$REPLY

Modified PATH without /usr/bin

2 Answers2