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Shellcheck #107

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Shellcheck #107

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5 changes: 3 additions & 2 deletions uninstall.sh
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Original file line number Diff line number Diff line change
Expand Up @@ -18,6 +18,7 @@ sleep 2

## now check what shell is running
shell_type=$(basename "$SHELL")
# shellcheck disable=SC2034
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What's the unused variable here?

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I think this is more that the specific variable is unknown, rather than undefined. The presence of ? would seem to indicate that the shell being used is not one of the known types and could be something else entirely.

Looking at the documentation - at least to my eyes - this looked like one of the exceptional conditions which produces a false warning, so as long as the script still works as intended, it can be ignored.

case "$shell_type" in
bash) startup_type="simple" ; startup_script="$HOME/.bashrc" ;;
zsh) startup_type="simple" ; startup_script="$HOME/.zshrc" ;;
Expand All @@ -33,7 +34,7 @@ esac
basher_keyword="basher5ea843"

echo ". following basher folders are in your path:"
echo $PATH | tr ':' "\n" | grep basher
echo "$PATH" | tr ':' "\n" | grep basher

if grep -q "$basher_keyword" "$startup_script" ; then
echo ". remove basher from startup script [$startup_script]"
Expand All @@ -44,7 +45,7 @@ if grep -q "$basher_keyword" "$startup_script" ; then
rm "$temp_file"
elif grep -q basher "$startup_script" ; then
grep basher "$startup_script"
die "Can't auto-remove the lines from $(basename $startup_script) - please do so manually "
die "Can't auto-remove the lines from $(basename "$startup_script") - please do so manually "
else
die "Can't find initialisation commands for basher"
fi
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