问题
Or more generally, how do I remove an item from a colon-separated list in a Bash environment variable?
I thought I had seen a simple way to do this years ago, using the more advanced forms of Bash variable expansion, but if so I\'ve lost track of it. A quick search of Google turned up surprisingly few relevant results and none that I would call \"simple\" or \"elegant\". For example, two methods using sed and awk, respectively:
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)
Does nothing straightforward exist? Is there anything analogous to a split() function in Bash?
Update:
It looks like I need to apologize for my intentionally-vague question; I was less interested in solving a specific use-case than in provoking good discussion. Fortunately, I got it!
There are some very clever techniques here. In the end, I\'ve added the following three functions to my toolbox. The magic happens in path_remove, which is based largely on Martin York\'s clever use of awk\'s RS variable.
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 != \"\'$1\'\"\' | sed \'s/:$//\'`; }
The only real cruft in there is the use of sed to remove the trailing colon. Considering how straightforward the rest of Martin\'s solution is, though, I\'m quite willing to live with it!
Related question: How do I manipulate $PATH elements in shell scripts?
回答1:
A minute with awk:
# Strip all paths with SDE in them.
#
export PATH=`echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}'`
Edit: It response to comments below:
$ 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 ${a} | awk -v RS=: -v ORS=: '/g/ {next} {print}'
/a/b/c/d/e:/a/b/c/d/f:
## Works fine: Again!
Edit in response to security problem: (that is not relevant to the question)
export PATH=$(echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}' | sed 's/:*$//')
This removes any trailing colons left by deleting the last entries, which would effectively add . to your path.
回答2:
My dirty hack:
echo ${PATH} > t1
vi t1
export PATH=$(cat t1)
回答3:
Since the big issue with substitution is the end cases, how about making the end cases no different to the other cases? If the path already had colons at the start and end, we could simply search for our desired string wrapped with colons. As it is, we can easily add those colons and remove them afterwards.
# 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
Pure bash :).
回答4:
Here's the simplest solution i can devise:
#!/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[*]}"
The above example will remove any element in $PATH that contains "usr". You can replace "*usr*" with "/home/user/bin" to remove just that element.
update per sschuberth
Even though i think spaces in a $PATH are a horrible idea, here's a solution that handles it:
PATH=$(IFS=':';t=($PATH);n=${#t[*]};a=();for ((i=0;i<n;i++)); do p="${t[i]%%*usr*}"; [ "${p}" ] && a[i]="${p}"; done;echo "${a[*]}");
or
IFS=':'
t=($PATH)
n=${#t[*]}
a=()
for ((i=0;i<n;i++)); do
p="${t[i]%%*usr*}"
[ "${p}" ] && a[i]="${p}"
done
echo "${a[*]}"
回答5:
Here's a one-liner that, despite the current accepted and highest rated answers, does not add invisible characters to PATH and can cope with paths that contain spaces:
export PATH=$(p=$(echo $PATH | tr ":" "\n" | grep -v "/cygwin/" | tr "\n" ":"); echo ${p%:})
Personally, I also find this easy to read / understand, and it only involves common commands instead of using awk.
回答6:
Here is a solution that:
- is pure Bash,
- does not invoke other processes (like 'sed' or 'awk'),
- does not change
IFS, - does not fork a sub-shell,
- handles paths with spaces, and
removes all occurrences of the argument in
PATH.removeFromPath() { local p d p=":$1:" d=":$PATH:" d=${d//$p/:} d=${d/#:/} PATH=${d/%:/} }
回答7:
function __path_remove(){
local D=":${PATH}:";
[ "${D/:$1:/:}" != "$D" ] && PATH="${D/:$1:/:}";
PATH="${PATH/#:/}";
export PATH="${PATH/%:/}";
}
Dug it out from my .bashrc file. When you play around with PATH, and it gets lost, awk/sed/grep becomes unavailable :-)
回答8:
The best pure bash option I have found so far is the following:
function path_remove {
PATH=${PATH/":$1"/} # delete any instances in the middle or at the end
PATH=${PATH/"$1:"/} # delete any instances at the beginning
}
This is based on the not quite correct answer to Add directory to $PATH if it's not already there over on Superuser.
回答9:
I've just been using the functions in the bash distribution, that have been there apparently since 1991. These are still in the bash-docs package on Fedora, and used to be used in /etc/profile, but no more...
$ 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;:\$;;"`
}
回答10:
I did write an answer to this here (using awk too). But i'm not sure that's what you are looking for? It at least looks clear to me what it does, instead of trying to fit into one line. For a simple one liner, though, that only removes stuff, i recommend
echo $PATH | tr ':' '\n' | awk '$0 != "/bin"' | paste -sd:
Replacing is
echo $PATH | tr ':' '\n' |
awk '$0 != "/bin"; $0 == "/bin" { print "/bar" }' | paste -sd:
or (shorter but less readable)
echo $PATH | tr ':' '\n' | awk '$0 == "/bin" { print "/bar"; next } 1' | paste -sd:
Anyway, for the same question, and a whole lot of useful answers, see here.
回答11:
Well, in bash, as it supports regular expression, I would simply do :
PATH=${PATH/:\/home\/user\/bin/}
回答12:
I like the three functions shown in @BenBlank's update to his original question. To generalize them, I use a 2-argument form, which allows me to set PATH or any other environment variable I want:
path_append () { path_remove $1 $2; export $1="${!1}:$2"; }
path_prepend () { path_remove $1 $2; export $1="$2:${!1}"; }
path_remove () { export $1="`echo -n ${!1} | awk -v RS=: -v ORS=: '$1 != "'$2'"' | sed 's/:$//'`"; }
Examples of use:
path_prepend PATH /usr/local/bin
path_append PERL5LIB "$DEVELOPMENT_HOME/p5/src/perlmods"
Note that I also added some quotation marks to allow for the proper processing of pathnames that contain spaces.
回答13:
What is the most elegant way to remove a path from the $PATH variable in Bash?
What's more elegant than awk?
path_remove () { export PATH=`echo -n $PATH | awk -v RS=: -v ORS=: '$0 != "'$1'"' | sed 's/:$//'`;
Python! It's a more readable and maintainable solution, and it is easy to inspect to see that it's really doing what you want.
Say you want to remove the first path element?
PATH="$(echo "$PATH" | python -c "import sys; path = sys.stdin.read().split(':'); del path[0]; print(':'.join(path))")"
(Instead of piping from echo, os.getenv['PATH'] would be a little shorter, and provided the same result as the above, but I'm worried that Python might do something with that environment variable, so it's probably best to pipe it directly from the environment you care about.)
Similarly to remove from the end:
PATH="$(echo "$PATH" | python -c "import sys; path = sys.stdin.read().split(':'); del path[-1]; print(':'.join(path))")"
To make these reusable shell functions that you can, for example, stick in your .bashrc file:
strip_path_first () {
PATH="$(echo "$PATH" |
python -c "import sys; path = sys.stdin.read().split(':'); del path[0]; print(':'.join(path))")"
}
strip_path_last () {
PATH="$(echo "$PATH" |
python -c "import sys; path = sys.stdin.read().split(':'); del path[-1]; print(':'.join(path))")"
}
回答14:
Yes, putting a colon at the end of PATH, for example, makes removing a path a bit less clumsy & error-prone.
path_remove () {
declare i newPATH
newPATH="${PATH}:"
for ((i=1; i<=${#@}; i++ )); do
#echo ${@:${i}:1}
newPATH="${newPATH//${@:${i}:1}:/}"
done
export PATH="${newPATH%:}"
return 0;
}
path_remove_all () {
declare i newPATH
shopt -s extglob
newPATH="${PATH}:"
for ((i=1; i<=${#@}; i++ )); do
newPATH="${newPATH//+(${@:${i}:1})*([^:]):/}"
#newPATH="${newPATH//+(${@:${i}:1})*([^:])+(:)/}"
done
shopt -u extglob
export PATH="${newPATH%:}"
return 0
}
path_remove /opt/local/bin /usr/local/bin
path_remove_all /opt/local /usr/local
回答15:
If you are concerned about removing duplicates in $PATH, the most elegant way, IMHO, would be not to add them in the first place. In 1 line:
if ! $( echo "$PATH" | tr ":" "\n" | grep -qx "$folder" ) ; then PATH=$PATH:$folder ; fi
$folder can be be replaced by anything, and may contain spaces ("/home/user/my documents")
回答16:
The most elegant pure bash solution I've found to date:
pathrm () {
local IFS=':'
local newpath
local dir
local pathvar=${2:-PATH}
for dir in ${!pathvar} ; do
if [ "$dir" != "$1" ] ; then
newpath=${newpath:+$newpath:}$dir
fi
done
export $pathvar="$newpath"
}
pathprepend () {
pathrm $1 $2
local pathvar=${2:-PATH}
export $pathvar="$1${!pathvar:+:${!pathvar}}"
}
pathappend () {
pathrm $1 $2
local pathvar=${2:-PATH}
export $pathvar="${!pathvar:+${!pathvar}:}$1"
}
回答17:
Most of the other suggested solutions rely only on string matching and don't take into account path segments containing special names like ., .., or ~. The bash function below resolves directory strings in its argument and in path segments to find logical directory matches as well as string matches.
rm_from_path() {
pattern="${1}"
dir=''
[ -d "${pattern}" ] && dir="$(cd ${pattern} && pwd)" # resolve to absolute path
new_path=''
IFS0=${IFS}
IFS=':'
for segment in ${PATH}; do
if [[ ${segment} == ${pattern} ]]; then # string match
continue
elif [[ -n ${dir} && -d ${segment} ]]; then
segment="$(cd ${segment} && pwd)" # resolve to absolute path
if [[ ${segment} == ${dir} ]]; then # logical directory match
continue
fi
fi
new_path="${new_path}${IFS}${segment}"
done
new_path="${new_path/#${IFS}/}" # remove leading colon, if any
IFS=${IFS0}
export PATH=${new_path}
}
Test:
$ mkdir -p ~/foo/bar/baz ~/foo/bar/bif ~/foo/boo/bang
$ PATH0=${PATH}
$ PATH=~/foo/bar/baz/.././../boo/././../bar:${PATH} # add dir with special names
$ rm_from_path ~/foo/boo/../bar/. # remove same dir with different special names
$ [ ${PATH} == ${PATH0} ] && echo 'PASS' || echo 'FAIL'
回答18:
Linux from Scratch defines three Bash functions in /etc/profile:
# Functions to help us manage paths. Second argument is the name of the
# path variable to be modified (default: PATH)
pathremove () {
local IFS=':'
local NEWPATH
local DIR
local PATHVARIABLE=${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
Ref: http://www.linuxfromscratch.org/blfs/view/svn/postlfs/profile.html
回答19:
Since this tends to be quite problematic, as in there IS NO elegant way, I recommend avoiding the problem by rearranging the solution: build your PATH up rather than attempt to tear it down.
I could be more specific if I knew your real problem context. In the interim, I will use a software build as the context.
A common problem with software builds is that it breaks on some machines, ultimately due to how someone has configured their default shell (PATH and other environment variables). The elegant solution is to make your build scripts immune by fully specifying the shell environment. Code your build scripts to set the PATH and other environment variables based on assembling pieces that you control, such as the location of the compiler, libraries, tools, components, etc. Make each configurable item something that you can individually set, verify, and then use appropriately in your script.
For example, I have a Maven-based WebLogic-targeted Java build that I inherited at my new employer. The build script is notorious for being fragile, and another new employee and I spent three weeks (not full time, just here and there, but still many hours) getting it to work on our machines. An essential step was that I took control of the PATH so that I knew exactly which Java, which Maven, and which WebLogic was being invoked. I created environment variables to point to each of those tools, then I calculated the PATH based on those plus a few others. Similar techniques tamed the other configurable settings, until we finally created a reproducible build.
By the way, don't use Maven, Java is okay, and only buy WebLogic if you absolutely need its clustering (but otherwise no, and especially not its proprietary features).
Best wishes.
回答20:
As with @litb, I contributed an answer to the question "How do I manipulate $PATH elements in shell scripts", so my main answer is there.
The 'split' functionality in bash and other Bourne shell derivatives is most neatly achieved with $IFS, the inter-field separator. For example, to set the positional arguments ($1, $2, ...) to the elements of PATH, use:
set -- $(IFS=":"; echo "$PATH")
It will work OK as long as there are no spaces in $PATH. Making it work for path elements containing spaces is a non-trivial exercise - left for the interested reader. It is probably simpler to deal with it using a scripting language such as Perl.
I also have a script, clnpath, which I use extensively for setting my PATH. I documented it in the answer to "How to keep from duplicating PATH variable in csh".
回答21:
What makes this problem annoying are the fencepost cases among first and last elements. The problem can be elegantly solved by changing IFS and using an array, but I don't know how to re-introduce the colon once the path is converted to array form.
Here is a slightly less elegant version that removes one directory from $PATH using string manipulation only. I have tested it.
#!/bin/bash
#
# remove_from_path dirname
#
# removes $1 from user's $PATH
if [ $# -ne 1 ]; then
echo "Usage: $0 pathname" 1>&2; exit 1;
fi
delendum="$1"
NEWPATH=
xxx="$IFS"
IFS=":"
for i in $PATH ; do
IFS="$xxx"
case "$i" in
"$delendum") ;; # do nothing
*) [ -z "$NEWPATH" ] && NEWPATH="$i" || NEWPATH="$NEWPATH:$i" ;;
esac
done
PATH="$NEWPATH"
echo "$PATH"
回答22:
Here's a Perl one-liner:
PATH=`perl -e '$a=shift;$_=$ENV{PATH};s#:$a(:)|^$a:|:$a$#$1#;print' /home/usr/bin`
The $a variable gets the path to be removed. The s (substitute) and print commands implicitly operate on the $_ variable.
回答23:
Good stuff here. I use this one to keep from adding dupes in the first place.
#!/bin/bash
#
######################################################################################
#
# Allows a list of additions to PATH with no dupes
#
# Patch code below into your $HOME/.bashrc file or where it
# will be seen at login.
#
# Can also be made executable and run as-is.
#
######################################################################################
# add2path=($HOME/bin .) ## uncomment space separated list
if [ $add2path ]; then ## skip if list empty or commented out
for nodup in ${add2path[*]}
do
case $PATH in ## case block thanks to MIKE511
$nodup:* | *:$nodup:* | *:$nodup ) ;; ## if found, do nothing
*) PATH=$PATH:$nodup ## else, add it to end of PATH or
esac ## *) PATH=$nodup:$PATH prepend to front
done
export PATH
fi
## debug add2path
echo
echo " PATH == $PATH"
echo
回答24:
With extended globbing enabled it's possible to do the following:
# delete all /opt/local paths in PATH
shopt -s extglob
printf "%s\n" "${PATH}" | tr ':' '\n' | nl
printf "%s\n" "${PATH//+(\/opt\/local\/)+([^:])?(:)/}" | tr ':' '\n' | nl
man bash | less -p extglob
回答25:
Extended globbing one-liner (well, sort of):
path_remove () { shopt -s extglob; PATH="${PATH//+(${1})+([^:])?(:)/}"; export PATH="${PATH%:}"; shopt -u extglob; return 0; }
There seems no need to escape slashes in $1.
path_remove () { shopt -s extglob; declare escArg="${1//\//\\/}"; PATH="${PATH//+(${escArg})+([^:])?(:)/}"; export PATH="${PATH%:}"; shopt -u extglob; return 0; }
回答26:
Adding colons to PATH we could also do something like:
path_remove () {
declare i newPATH
# put a colon at the beginning & end AND double each colon in-between
newPATH=":${PATH//:/::}:"
for ((i=1; i<=${#@}; i++)); do
#echo ${@:${i}:1}
newPATH="${newPATH//:${@:${i}:1}:/}" # s/:\/fullpath://g
done
newPATH="${newPATH//::/:}"
newPATH="${newPATH#:}" # remove leading colon
newPATH="${newPATH%:}" # remove trailing colon
unset PATH
PATH="${newPATH}"
export PATH
return 0
}
path_remove_all () {
declare i newPATH extglobVar
extglobVar=0
# enable extended globbing if necessary
[[ ! $(shopt -q extglob) ]] && { shopt -s extglob; extglobVar=1; }
newPATH=":${PATH}:"
for ((i=1; i<=${#@}; i++ )); do
newPATH="${newPATH//:+(${@:${i}:1})*([^:])/}" # s/:\/path[^:]*//g
done
newPATH="${newPATH#:}" # remove leading colon
newPATH="${newPATH%:}" # remove trailing colon
# disable extended globbing if it was enabled in this function
[[ $extglobVar -eq 1 ]] && shopt -u extglob
unset PATH
PATH="${newPATH}"
export PATH
return 0
}
path_remove /opt/local/bin /usr/local/bin
path_remove_all /opt/local /usr/local
回答27:
In path_remove_all (by proxxy):
-newPATH="${newPATH//:+(${@:${i}:1})*([^:])/}"
+newPATH="${newPATH//:${@:${i}:1}*([^:])/}" # s/:\/path[^:]*//g
回答28:
While this is a very old thread, I thought this solution might be of interest:
PATH="/usr/lib/ccache:/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games"
REMOVE="ccache" # whole or part of a path :)
export PATH=$(IFS=':';p=($PATH);unset IFS;p=(${p[@]%%$REMOVE});IFS=':';echo "${p[*]}";unset IFS)
echo $PATH # outputs /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games
found it on this blog post. I think I like this one most :)
回答29:
I took a slightly different approach than most people here and focussed specifically on just string manipulation, like so:
path_remove () {
if [[ ":$PATH:" == *":$1:"* ]]; then
local dirs=":$PATH:"
dirs=${dirs/:$1:/:}
export PATH="$(__path_clean $dirs)"
fi
}
__path_clean () {
local dirs=${1%?}
echo ${dirs#?}
}
The above is a simplified example of the final functions I use. I've also created path_add_before and path_add_after allowing you insert a path before/after a specified path already in PATH.
The full set of functions are available in path_helpers.sh in my dotfiles. They fully support removing/appending/prepending/inserting at beginning/middle/end of the PATH string.
回答30:
The trailing ':' is caused by the fact that you are setting the line ending, not the separator. I use resource limitted units and like to pack everything into a single script, without these oddities:
path_remove () {
PATH="$(echo -n $PATH | awk -v RS=: -v ORS= '$0 != "'$1'"{print s _ $0;s=":"}')"
}
来源：https://stackoverflow.com/questions/370047/what-is-the-most-elegant-way-to-remove-a-path-from-the-path-variable-in-bash