求助,为何bash里用echo $PATH得到的PATH怎么这么复杂
redhat吧
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dayback 楼主
这是怎么一回事啊?是不是因为多用户的原因吗?我发现即使我把~/.bashrc删除了,PATH还是这么复杂。
求助~
2017年07月21日 05点07分 1
level 12
/etc/profile 呢
2017年07月30日 13点07分 2
level 1
dayback 楼主
pathmunge () {
case ":${PATH}:" in
*:"$1":*)
;;
*)
if [ "$2" = "after" ] ; then
PATH=$PATH:$1
else
PATH=$1:$PATH
fi
esac
}
if [ -x /usr/bin/id ]; then
if [ -z "$EUID" ]; then
# ksh workaround
EUID=`/usr/bin/id -u`
UID=`/usr/bin/id -ru`
fi
USER="`/usr/bin/id -un`"
LOGNAME=$USER
MAIL="/var/spool/mail/$USER"
fi
# Path manipulation
if [ "$EUID" = "0" ]; then
pathmunge /sbin
pathmunge /usr/sbin
pathmunge /usr/local/sbin
else
pathmunge /usr/local/sbin after
pathmunge /usr/sbin after
pathmunge /sbin after
fi
HOSTNAME=`/bin/hostname 2>/dev/null`
HISTSIZE=1000
if [ "$HISTCONTROL" = "ignorespace" ] ; then
export HISTCONTROL=ignoreboth
else
export HISTCONTROL=ignoredups
fi
export PATH USER LOGNAME MAIL HOSTNAME HISTSIZE HISTCONTROL
# By default, we want umask to get set. This sets it for login shell
# Current threshold for system reserved uid/gids is 200
# You could check uidgid reservation validity in
# /usr/share/doc/setup-*/uidgid file
if [ $UID -gt 199 ] && [ "`/usr/bin/id -gn`" = "`/usr/bin/id -un`" ]; then
umask 002
else
umask 022
fi
for i in /etc/profile.d/*.sh ; do
if [ -r "$i" ]; then
if [ "${-#*i}" != "$-" ]; then
. "$i"
else
. "$i" >/dev/null 2>&1
fi
fi
done
unset i
unset -f pathmunge
=============
是不是哪儿有循环了?
2017年08月21日 09点08分 3
1