从Bash中的$ PATH变量中删除路径的最优雅方法是什么？

Question

或者更一般地说,如何从Bash环境变量中以冒号分隔的列表中删除项？

我想我多年前已经看到了一种简单的方法,使用更高级的Bash变量扩展形式,但如果是这样的话,我已经忘记了它.对谷歌的快速搜索出乎意料地少了几个相关结果,没有一个我称之为"简单"或"优雅".例如,分别使用sed和awk的两种方法:

PATH=$(echo $PATH | sed -e 's;:\?/home/user/bin;;' -e 's;/home/user/bin:\?;;')
PATH=!(awk -F: '{for(i=1;i<=NF;i++){if(!($i in a)){a[$i];printf s$i;s=":"}}}'<<<$PATH)

什么都不直接存在？有什么类似于Bash中的split()函数吗？

更新:
看起来我需要为我的故意模糊的问题道歉; 我对解决特定用例的兴趣不如激发良好的讨论.幸运的是,我明白了!

这里有一些非常聪明的技巧.最后,我在工具箱中添加了以下三个功能.魔法发生在path_remove中,这主要基于Martin York巧妙使用awk的RS变量.

path_append ()  { path_remove $1; export PATH="$PATH:$1"; }
path_prepend () { path_remove $1; export PATH="$1:$PATH"; }
path_remove ()  { export PATH=`echo -n $PATH | awk -v RS=: -v ORS=: '$0 != "'$1'"' | sed 's/:$//'`; }

唯一真正的缺点是使用sed去除尾部结肠.考虑到马丁的其他解决方案是多么简单,但我非常愿意接受它!

---

相关问题: 如何在shell脚本中操作$ PATH元素？

Answer 1

我的肮脏黑客:

echo ${PATH} > t1
vi t1
export PATH=$(cat t1)

Answer 2

与awk一分钟:

# Strip all paths with SDE in them.
#
export PATH=`echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}'`

编辑:它回复以下评论:

$ export a="/a/b/c/d/e:/a/b/c/d/g/k/i:/a/b/c/d/f:/a/b/c/g:/a/b/c/d/g/i"
$ echo ${a}
/a/b/c/d/e:/a/b/c/d/f:/a/b/c/g:/a/b/c/d/g/i

## Remove multiple (any directory with a: all of them)
$ echo ${a} | awk -v RS=: -v ORS=: '/a/ {next} {print}'
## Works fine all removed

## Remove multiple including last two: (any directory with g)
$ echo ${a} | awk -v RS=: -v ORS=: '/g/ {next} {print}'
/a/b/c/d/e:/a/b/c/d/f:
## Works fine: Again!

编辑以响应安全问题:(与问题无关)

export PATH=$(echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}' | sed 's/:*$//')

这将通过删除最后一个条目来删除任何尾随冒号,这将有效地添加.到您的路径中.

Answer 3

由于替换的最大问题是最终案例,如何使最终案例与其他案例没有什么不同？如果路径在开始和结束时已经有冒号,我们可以简单地搜索用冒号包裹的所需字符串.事实上,我们可以轻松添加这些冒号并在之后删除它们.

# PATH => /bin:/opt/a dir/bin:/sbin
WORK=:$PATH:
# WORK => :/bin:/opt/a dir/bin:/sbin:
REMOVE='/opt/a dir/bin'
WORK=${WORK/:$REMOVE:/:}
# WORK => :/bin:/sbin:
WORK=${WORK%:}
WORK=${WORK#:}
PATH=$WORK
# PATH => /bin:/sbin

纯粹的打击:).

Answer 4

这是我可以设计的最简单的解决方案:

#!/bin/bash
IFS=:
# convert it to an array
t=($PATH)
unset IFS
# perform any array operations to remove elements from the array
t=(${t[@]%%*usr*})
IFS=:
# output the new array
echo "${t[*]}"

上面的示例将删除$ PATH中包含"usr"的任何元素.您可以将"*usr*"替换为"/ home/user/bin"以仅删除该元素.

每个sschuberth 更新

即使我认为空间中的空间$PATH是一个可怕的想法,这里有一个处理它的解决方案:

PATH=$(IFS=':';t=($PATH);n=${#t[*]};a=();for ((i=0;i<n;i++)); do p="${t[i]%%*usr*}"; [ "${p}" ] && a[i]="${p}"; done;echo "${a[*]}");

要么

IFS=':'
t=($PATH)
n=${#t[*]}
a=()
for ((i=0;i<n;i++)); do
  p="${t[i]%%*usr*}"
  [ "${p}" ] && a[i]="${p}"
done
echo "${a[*]}"

Answer 5

这是一个单线程,尽管当前接受和评级最高的答案,但不会向PATH添加不可见的字符,并且可以处理包含空格的路径:

export PATH=$(p=$(echo $PATH | tr ":" "\n" | grep -v "/cygwin/" | tr "\n" ":"); echo ${p%:})

就个人而言,我也发现这很容易阅读/理解,它只涉及常见命令而不是使用awk.

Answer 6

这是一个解决方案:

• 是纯粹的Bash,
• 不会调用其他进程(如'sed'或'awk'),
• 不改变IFS,
• 不分叉子壳,
• 处理带空格的路径,和

• 删除所有出现的参数PATH.

  removeFromPath() {
     local p d
     p=":$1:"
     d=":$PATH:"
     d=${d//$p/:}
     d=${d/#:/}
     PATH=${d/%:/}
  }

Answer 7

function __path_remove(){
local D =":$ {PATH}:";
["$ {D /:$ 1:/:}"!="$ D"] && PATH ="$ {D /:$ 1:/:}";
PATH = "$ {PATH /#:/}";
export PATH ="$ {PATH /%:/}";
}

从我的.bashrc文件中挖出来.当你玩PATH时,它会丢失,awk/sed/grep变得不可用:-)

Answer 8

到目前为止我发现的最好的纯bash选项如下:

function path_remove {
  PATH=${PATH/":$1"/} # delete any instances in the middle or at the end
  PATH=${PATH/"$1:"/} # delete any instances at the beginning
}

这是基于添加目录到$ PATH的不太正确的答案,如果它还没有超级用户.

Answer 9

我一直在使用bash发行版中的函数,这些函数自1991年以来一直存在.这些函数仍然在Fedora上的bash-docs包中,过去常用/etc/profile,但不再...

$ rpm -ql bash-doc |grep pathfunc
/usr/share/doc/bash-4.2.20/examples/functions/pathfuncs
$ cat $(!!)
cat $(rpm -ql bash-doc |grep pathfunc)
#From: "Simon J. Gerraty" <sjg@zen.void.oz.au>
#Message-Id: <199510091130.VAA01188@zen.void.oz.au>
#Subject: Re: a shell idea?
#Date: Mon, 09 Oct 1995 21:30:20 +1000


# NAME:
#       add_path.sh - add dir to path
#
# DESCRIPTION:
#       These functions originated in /etc/profile and ksh.kshrc, but
#       are more useful in a separate file.
#
# SEE ALSO:
#       /etc/profile
#
# AUTHOR:
#       Simon J. Gerraty <sjg@zen.void.oz.au>

#       @(#)Copyright (c) 1991 Simon J. Gerraty
#
#       This file is provided in the hope that it will
#       be of use.  There is absolutely NO WARRANTY.
#       Permission to copy, redistribute or otherwise
#       use this file is hereby granted provided that
#       the above copyright notice and this notice are
#       left intact.

# is $1 missing from $2 (or PATH) ?
no_path() {
        eval "case :\$${2-PATH}: in *:$1:*) return 1;; *) return 0;; esac"
}
# if $1 exists and is not in path, append it
add_path () {
  [ -d ${1:-.} ] && no_path $* && eval ${2:-PATH}="\$${2:-PATH}:$1"
}
# if $1 exists and is not in path, prepend it
pre_path () {
  [ -d ${1:-.} ] && no_path $* && eval ${2:-PATH}="$1:\$${2:-PATH}"
}
# if $1 is in path, remove it
del_path () {
  no_path $* || eval ${2:-PATH}=`eval echo :'$'${2:-PATH}: |
    sed -e "s;:$1:;:;g" -e "s;^:;;" -e "s;:\$;;"`
}

Answer 10

Linux from Scratch 在 中定义了三个 Bash 函数/etc/profile：

# Functions to help us manage paths.  Second argument is the name of the
# path variable to be modified (default: PATH)
pathremove () {
        local IFS=':'
        local NEWPATH
        local DIR
        local PATHVARIABLE=${2:-PATH}
        for DIR in ${!PATHVARIABLE} ; do
                if [ "$DIR" != "$1" ] ; then
                  NEWPATH=${NEWPATH:+$NEWPATH:}$DIR
                fi
        done
        export $PATHVARIABLE="$NEWPATH"
}

pathprepend () {
        pathremove $1 $2
        local PATHVARIABLE=${2:-PATH}
        export $PATHVARIABLE="$1${!PATHVARIABLE:+:${!PATHVARIABLE}}"
}

pathappend () {
        pathremove $1 $2
        local PATHVARIABLE=${2:-PATH}
        export $PATHVARIABLE="${!PATHVARIABLE:+${!PATHVARIABLE}:}$1"
}

export -f pathremove pathprepend pathappend

参考： http: //www.linuxfromscratch.org/blfs/view/svn/postlfs/profile.html