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Fix bash script without conditional function

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This commit is contained in:
Dan
2024-12-29 18:13:50 +00:00
parent 6508fec945
commit b63a9cd2e0
1 changed files with 21 additions and 19 deletions
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40
src/shell-integration/bash/ghostty.bash
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@ -6,6 +6,26 @@
# my simple bash usage this is working. If a bash expert wants to
# improve this please do!
sudo_function() {
builtin local sudo_has_sudoedit_flags="no"
for arg in "$@"; do
# Check if argument is '-e' or '--edit' (sudoedit flags)
if [[ "$arg" == "-e" || $arg == "--edit" ]]; then
sudo_has_sudoedit_flags="yes"
builtin break
fi
# Check if argument is neither an option nor a key-value pair
if [[ "$arg" != -* && "$arg" != *=* ]]; then
builtin break
fi
done
if [[ "$sudo_has_sudoedit_flags" == "yes" ]]; then
builtin command sudo "$@";
else
builtin command sudo TERMINFO="$TERMINFO" "$@";
fi
}
# We need to be in interactive mode and we need to have the Ghostty
# resources dir set which also tells us we're running in Ghostty.
if [[ "$-" != *i* ]] ; then builtin return; fi
@ -117,25 +137,7 @@ function __ghostty_precmd() {
if [[ "$GHOSTTY_SHELL_INTEGRATION_NO_SUDO" != "1" ]] && [[ -n "$TERMINFO" ]]; then
# Wrap `sudo` command to ensure Ghostty terminfo is preserved
# shellcheck disable=SC2317
sudo() {
builtin local sudo_has_sudoedit_flags="no"
for arg in "$@"; do
# Check if argument is '-e' or '--edit' (sudoedit flags)
if [[ "$arg" == "-e" || $arg == "--edit" ]]; then
sudo_has_sudoedit_flags="yes"
builtin break
fi
# Check if argument is neither an option nor a key-value pair
if [[ "$arg" != -* && "$arg" != *=* ]]; then
builtin break
fi
done
if [[ "$sudo_has_sudoedit_flags" == "yes" ]]; then
builtin command sudo "$@";
else
builtin command sudo TERMINFO="$TERMINFO" "$@";
fi
}
alias sudo="sudo_function"
fi
if [[ "$GHOSTTY_SHELL_INTEGRATION_NO_TITLE" != 1 ]]; then