是否有一种从PATH类shell变量中删除元素的惯用方法？
那是我想要的
PATH=/home/joe/bin:/usr/local/bin:/usr/bin:/bin:/path/to/app/bin:.
和删除或替换的/path/to/app/bin,而不重挫可变的其余部分.允许我将新元素放在任意位置的额外点数.目标将由明确定义的字符串识别,并且可以在列表中的任何点处发生.
我知道我已经看到了这一点,并且可以自己拼凑一些东西,但我正在寻找一个很好的方法.便携性和标准化加分.
我使用bash,但欢迎你最喜欢的shell中的例子.
这里的上下文是需要在多个版本之间方便地切换(一个用于进行分析,另一个用于处理框架)的大型科学分析包,它产生几十个可执行文件,数据存储在文件系统周围,并使用环境变量帮助找到所有这些东西.我想编写一个选择版本的脚本,并且需要能够删除$PATH与当前活动版本相关的元素,并用与新版本相关的相同元素替换它们.
这与$PATH在重新运行登录脚本等时防止重复元素的问题有关.
以前的类似问题:如何避免在csh中复制路径变量
随后的类似问题:从Bash中的$ PATH变量中删除路径的最优雅方法是什么？
Jonathan Lef.. 19
从dmckee处理建议的解决方案:
某些版本的Bash可能允许在函数名称中使用连字符,而其他版本(MacOS X)则不允许.
我觉得在函数结束之前不需要立即使用return.
我不认为需要所有的冒号.
我不明白为什么你有逐个元素的路径输出值.可以认为export等同于设置(甚至创建)全局变量 - 尽可能避免这种情况.
我不确定你的期望replace-path PATH $PATH /usr是什么,但它没有达到我的预期.
考虑一个开始时包含的PATH值:
. /Users/jleffler/bin /usr/local/postgresql/bin /usr/local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /usr/local/bin /usr/bin /bin /sw/bin /usr/sbin /sbin
我得到的结果(来自' replace-path PATH $PATH /usr')是:
. /Users/jleffler/bin /local/postgresql/bin /local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /local/bin /bin /bin /sw/bin /sbin /sbin
我本来希望得到我的原始路径,因为/ usr不会显示为(完整的)路径元素,仅作为路径元素的一部分.
这可以replace-path通过修改其中一个sed命令来修复:
export $path=$(echo -n $list | tr ":" "\n" | sed "s:^$removestr\$:$replacestr:" |
tr "\n" ":" | sed "s|::|:|g")
我使用':'而不是'|' 从'|'开始分离替代品的部分 可能(理论上)出现在路径组件中,而根据PATH的定义,冒号不能.我观察到第二个sed可以从PATH中间消除当前目录.也就是说,PATH的合法(虽然有悖常理)值可能是:
PATH=/bin::/usr/local/bin
处理完后,当前目录将不再位于PATH上.
锚定匹配的类似更改适用于path-element-by-pattern:
export $target=$(echo -n $list | tr ":" "\n" | grep -m 1 "^$pat\$")
我注意到grep -m 1它不是标准的(它是一个GNU扩展,也可以在MacOS X上使用).事实上,-n选择权echo也是非标准的; 你最好只删除由于将换行符从echo转换为冒号而添加的尾部冒号.由于path-element-by-pattern只使用一次,因此会产生不良的副作用(它会破坏任何预先存在的导出变量$removestr),它可以被它的主体明智地替换掉.这一点,以及更自由地使用引号来避免空格或不需要的文件名扩展问题,导致:
# path_tools.bash # # A set of tools for manipulating ":" separated lists like the # canonical $PATH variable. # # /bin/sh compatibility can probably be regained by replacing $( ) # style command expansion with ` ` style ############################################################################### # Usage: # # To remove a path: # replace_path PATH $PATH /exact/path/to/remove # replace_path_pattern PATH $PATH# # To replace a path: # replace_path PATH $PATH /exact/path/to/remove /replacement/path # replace_path_pattern PATH $PATH /replacement/path # ############################################################################### # Remove or replace an element of $1 # # $1 name of the shell variable to set (e.g. PATH) # $2 a ":" delimited list to work from (e.g. $PATH) # $3 the precise string to be removed/replaced # $4 the replacement string (use "" for removal) function replace_path () { path=$1 list=$2 remove=$3 replace=$4 # Allowed to be empty or unset export $path=$(echo "$list" | tr ":" "\n" | sed "s:^$remove\$:$replace:" | tr "\n" ":" | sed 's|:$||') } # Remove or replace an element of $1 # # $1 name of the shell variable to set (e.g. PATH) # $2 a ":" delimited list to work from (e.g. $PATH) # $3 a grep pattern identifying the element to be removed/replaced # $4 the replacement string (use "" for removal) function replace_path_pattern () { path=$1 list=$2 removepat=$3 replacestr=$4 # Allowed to be empty or unset removestr=$(echo "$list" | tr ":" "\n" | grep -m 1 "^$removepat\$") replace_path "$path" "$list" "$removestr" "$replacestr" }
我有一个Perl脚本调用echopath,在调试类似PATH的变量时我发现它很有用:
#!/usr/bin/perl -w
#
# "@(#)$Id: echopath.pl,v 1.7 1998/09/15 03:16:36 jleffler Exp $"
#
# Print the components of a PATH variable one per line.
# If there are no colons in the arguments, assume that they are
# the names of environment variables.
@ARGV = $ENV{PATH} unless @ARGV;
foreach $arg (@ARGV)
{
$var = $arg;
$var = $ENV{$arg} if $arg =~ /^[A-Za-z_][A-Za-z_0-9]*$/;
$var = $arg unless $var;
@lst = split /:/, $var;
foreach $val (@lst)
{
print "$val\n";
}
}
当我在下面的测试代码上运行修改后的解决方案时:
echo xpath=$PATH replace_path xpath $xpath /usr echopath $xpath echo xpath=$PATH replace_path_pattern xpath $xpath /usr/bin /work/bin echopath xpath echo xpath=$PATH replace_path_pattern xpath $xpath "/usr/.*/bin" /work/bin echopath xpath
输出是:
. /Users/jleffler/bin /usr/local/postgresql/bin /usr/local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /usr/local/bin /usr/bin /bin /sw/bin /usr/sbin /sbin . /Users/jleffler/bin /usr/local/postgresql/bin /usr/local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /usr/local/bin /work/bin /bin /sw/bin /usr/sbin /sbin . /Users/jleffler/bin /work/bin /usr/local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /usr/local/bin /usr/bin /bin /sw/bin /usr/sbin /sbin
这对我来说是正确的 - 至少,我对问题的定义是正确的.
我注意到echopath LD_LIBRARY_PATH评估$LD_LIBRARY_PATH.如果你的功能能够做到这一点会很好,所以用户可以输入:
replace_path PATH /usr/bin /work/bin
这可以通过使用:
list=$(eval echo '$'$path)
这导致代码的这次修订:
# path_tools.bash # # A set of tools for manipulating ":" separated lists like the # canonical $PATH variable. # # /bin/sh compatibility can probably be regained by replacing $( ) # style command expansion with ` ` style ############################################################################### # Usage: # # To remove a path: # replace_path PATH /exact/path/to/remove # replace_path_pattern PATH# # To replace a path: # replace_path PATH /exact/path/to/remove /replacement/path # replace_path_pattern PATH /replacement/path # ############################################################################### # Remove or replace an element of $1 # # $1 name of the shell variable to set (e.g. PATH) # $2 the precise string to be removed/replaced # $3 the replacement string (use "" for removal) function replace_path () { path=$1 list=$(eval echo '$'$path) remove=$2 replace=$3 # Allowed to be empty or unset export $path=$(echo "$list" | tr ":" "\n" | sed "s:^$remove\$:$replace:" | tr "\n" ":" | sed 's|:$||') } # Remove or replace an element of $1 # # $1 name of the shell variable to set (e.g. PATH) # $2 a grep pattern identifying the element to be removed/replaced # $3 the replacement string (use "" for removal) function replace_path_pattern () { path=$1 list=$(eval echo '$'$path) removepat=$2 replacestr=$3 # Allowed to be empty or unset removestr=$(echo "$list" | tr ":" "\n" | grep -m 1 "^$removepat\$") replace_path "$path" "$removestr" "$replacestr" }
以下修订测试现在也适用:
echo xpath=$PATH replace_path xpath /usr echopath xpath echo xpath=$PATH replace_path_pattern xpath /usr/bin /work/bin echopath xpath echo xpath=$PATH replace_path_pattern xpath "/usr/.*/bin" /work/bin echopath xpath
它产生与以前相同的输出.
从dmckee处理建议的解决方案:
某些版本的Bash可能允许在函数名称中使用连字符,而其他版本(MacOS X)则不允许.
我觉得在函数结束之前不需要立即使用return.
我不认为需要所有的冒号.
我不明白为什么你有逐个元素的路径输出值.可以认为export等同于设置(甚至创建)全局变量 - 尽可能避免这种情况.
我不确定你的期望replace-path PATH $PATH /usr是什么,但它没有达到我的预期.
考虑一个开始时包含的PATH值:
. /Users/jleffler/bin /usr/local/postgresql/bin /usr/local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /usr/local/bin /usr/bin /bin /sw/bin /usr/sbin /sbin
我得到的结果(来自' replace-path PATH $PATH /usr')是:
. /Users/jleffler/bin /local/postgresql/bin /local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /local/bin /bin /bin /sw/bin /sbin /sbin
我本来希望得到我的原始路径,因为/ usr不会显示为(完整的)路径元素,仅作为路径元素的一部分.
这可以replace-path通过修改其中一个sed命令来修复:
export $path=$(echo -n $list | tr ":" "\n" | sed "s:^$removestr\$:$replacestr:" |
tr "\n" ":" | sed "s|::|:|g")
我使用':'而不是'|' 从'|'开始分离替代品的部分 可能(理论上)出现在路径组件中,而根据PATH的定义,冒号不能.我观察到第二个sed可以从PATH中间消除当前目录.也就是说,PATH的合法(虽然有悖常理)值可能是:
PATH=/bin::/usr/local/bin
处理完后,当前目录将不再位于PATH上.
锚定匹配的类似更改适用于path-element-by-pattern:
export $target=$(echo -n $list | tr ":" "\n" | grep -m 1 "^$pat\$")
我注意到grep -m 1它不是标准的(它是一个GNU扩展,也可以在MacOS X上使用).事实上,-n选择权echo也是非标准的; 你最好只删除由于将换行符从echo转换为冒号而添加的尾部冒号.由于path-element-by-pattern只使用一次,因此会产生不良的副作用(它会破坏任何预先存在的导出变量$removestr),它可以被它的主体明智地替换掉.这一点,以及更自由地使用引号来避免空格或不需要的文件名扩展问题,导致:
# path_tools.bash # # A set of tools for manipulating ":" separated lists like the # canonical $PATH variable. # # /bin/sh compatibility can probably be regained by replacing $( ) # style command expansion with ` ` style ############################################################################### # Usage: # # To remove a path: # replace_path PATH $PATH /exact/path/to/remove # replace_path_pattern PATH $PATH# # To replace a path: # replace_path PATH $PATH /exact/path/to/remove /replacement/path # replace_path_pattern PATH $PATH /replacement/path # ############################################################################### # Remove or replace an element of $1 # # $1 name of the shell variable to set (e.g. PATH) # $2 a ":" delimited list to work from (e.g. $PATH) # $3 the precise string to be removed/replaced # $4 the replacement string (use "" for removal) function replace_path () { path=$1 list=$2 remove=$3 replace=$4 # Allowed to be empty or unset export $path=$(echo "$list" | tr ":" "\n" | sed "s:^$remove\$:$replace:" | tr "\n" ":" | sed 's|:$||') } # Remove or replace an element of $1 # # $1 name of the shell variable to set (e.g. PATH) # $2 a ":" delimited list to work from (e.g. $PATH) # $3 a grep pattern identifying the element to be removed/replaced # $4 the replacement string (use "" for removal) function replace_path_pattern () { path=$1 list=$2 removepat=$3 replacestr=$4 # Allowed to be empty or unset removestr=$(echo "$list" | tr ":" "\n" | grep -m 1 "^$removepat\$") replace_path "$path" "$list" "$removestr" "$replacestr" }
我有一个Perl脚本调用echopath,在调试类似PATH的变量时我发现它很有用:
#!/usr/bin/perl -w
#
# "@(#)$Id: echopath.pl,v 1.7 1998/09/15 03:16:36 jleffler Exp $"
#
# Print the components of a PATH variable one per line.
# If there are no colons in the arguments, assume that they are
# the names of environment variables.
@ARGV = $ENV{PATH} unless @ARGV;
foreach $arg (@ARGV)
{
$var = $arg;
$var = $ENV{$arg} if $arg =~ /^[A-Za-z_][A-Za-z_0-9]*$/;
$var = $arg unless $var;
@lst = split /:/, $var;
foreach $val (@lst)
{
print "$val\n";
}
}
当我在下面的测试代码上运行修改后的解决方案时:
echo xpath=$PATH replace_path xpath $xpath /usr echopath $xpath echo xpath=$PATH replace_path_pattern xpath $xpath /usr/bin /work/bin echopath xpath echo xpath=$PATH replace_path_pattern xpath $xpath "/usr/.*/bin" /work/bin echopath xpath
输出是:
. /Users/jleffler/bin /usr/local/postgresql/bin /usr/local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /usr/local/bin /usr/bin /bin /sw/bin /usr/sbin /sbin . /Users/jleffler/bin /usr/local/postgresql/bin /usr/local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /usr/local/bin /work/bin /bin /sw/bin /usr/sbin /sbin . /Users/jleffler/bin /work/bin /usr/local/mysql/bin /Users/jleffler/perl/v5.10.0/bin /usr/local/bin /usr/bin /bin /sw/bin /usr/sbin /sbin
这对我来说是正确的 - 至少,我对问题的定义是正确的.
我注意到echopath LD_LIBRARY_PATH评估$LD_LIBRARY_PATH.如果你的功能能够做到这一点会很好,所以用户可以输入:
replace_path PATH /usr/bin /work/bin
这可以通过使用:
list=$(eval echo '$'$path)
这导致代码的这次修订:
# path_tools.bash # # A set of tools for manipulating ":" separated lists like the # canonical $PATH variable. # # /bin/sh compatibility can probably be regained by replacing $( ) # style command expansion with ` ` style ############################################################################### # Usage: # # To remove a path: # replace_path PATH /exact/path/to/remove # replace_path_pattern PATH# # To replace a path: # replace_path PATH /exact/path/to/remove /replacement/path # replace_path_pattern PATH /replacement/path # ############################################################################### # Remove or replace an element of $1 # # $1 name of the shell variable to set (e.g. PATH) # $2 the precise string to be removed/replaced # $3 the replacement string (use "" for removal) function replace_path () { path=$1 list=$(eval echo '$'$path) remove=$2 replace=$3 # Allowed to be empty or unset export $path=$(echo "$list" | tr ":" "\n" | sed "s:^$remove\$:$replace:" | tr "\n" ":" | sed 's|:$||') } # Remove or replace an element of $1 # # $1 name of the shell variable to set (e.g. PATH) # $2 a grep pattern identifying the element to be removed/replaced # $3 the replacement string (use "" for removal) function replace_path_pattern () { path=$1 list=$(eval echo '$'$path) removepat=$2 replacestr=$3 # Allowed to be empty or unset removestr=$(echo "$list" | tr ":" "\n" | grep -m 1 "^$removepat\$") replace_path "$path" "$removestr" "$replacestr" }
以下修订测试现在也适用:
echo xpath=$PATH replace_path xpath /usr echopath xpath echo xpath=$PATH replace_path_pattern xpath /usr/bin /work/bin echopath xpath echo xpath=$PATH replace_path_pattern xpath "/usr/.*/bin" /work/bin echopath xpath
它产生与以前相同的输出.
重新发布我的答案什么是从Bash中的$ PATH变量中删除路径的最优雅方法？:
#!/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 "${t[*]}"
或者单行:
PATH=$(IFS=':';t=($PATH);unset IFS;t=(${t[@]%%*usr*});IFS=':';echo "${t[*]}");
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