创建带参数的bash函数

Question

我想在输入rm -rf到终端时提示确认问题，如果回答"是"则执行操作。我在直接rm命令中找到了这样的内容:

rm() {
read -p "Type password? " pass
if [ "$pass" = "1234" ]; then
command rm "$@"
else
#Termination and echo a message wrong pass. 
fi
}

用于添加到。bashrc。并将其转换为:

rm(){
if [ $1 -eq "-" ] && [ $2 -eq "r" ] && [ $2 -eq "r" ]; then
read -p "Do you really want to delete?" ans
if [ ans -eq "yes" ]; then
rm -rf $@;
else
fi

else
fi
}

但我知道它不会工作。那么我如何检测rm -rf命令并提示我的自定义确认问题?

编辑:我现在有这个，但仍然没有运气:

rm(){
if [ $1 = "-rf" ]; then
read -p "Do you really want to delete?" ans;

if [ $ans = "yes" ]; then
rm -rf "$@";
fi
fi
}

Answer 1

如果您不介意rm -fr,rm -r -f等不被捕获，这将工作得很好:

rm()
if [[ $1 = -rf ]]; then
read -p 'Are you sure? ' reply
if [[ $reply = yes ]]; then
command rm "$@"
fi
fi

Answer 2

让我们看看。要涵盖所有情况(-rf,-fr,-r -f,--recursive --force,-R --force;或其他组合，例如-vrvfv-getopts可能有帮助)。

这里有一个函数可以让你开始。

rm() {
local use_force=0
local is_recursive=0
for arg; do
case "$arg" in
--force) use_force=1 ;;
--recursive) is_recursive=1 ;;
# skip remaining long options:
--*) continue ;;
# handle short options:
-*)
while getopts :frFR flag "$@"; do
case "$flag" in
[fF]) use_force=1 ;;
[rR]) is_recursive=1 ;;
esac
done
;;
esac
done
if [ "$use_force" = 1 ] && [ "$is_recursive" = 1 ]; then
read -p 'Do you really want to delete?' answer
case "$answer" in
[yY]*) command rm "$@" ;;
*) echo 'Skipping delete ...' ;;
esac
else
# only -f/--force, only -R/-r/--recursive, or neither
command rm "$@"
fi
}

你没有真正指定你想如何处理rm -r dir或rm -f file(所以只有递归或只有强制，而不是组合)。我假定在这种情况下您不会收到确认问题。