如何在外壳程序脚本中操作$ PATH元素? [英] How do I manipulate $PATH elements in shell scripts?
问题描述
是否有一种惯用的方法从类似PATH的shell变量中删除元素?
Is there a idiomatic way of removing elements from PATH-like shell variables?
那是我想要的
PATH=/home/joe/bin:/usr/local/bin:/usr/bin:/bin:/path/to/app/bin:.
并删除或替换 /path/to/app/bin
,而不会破坏其余的变量.允许我在任意位置放入新元素的加分.目标可以通过定义良好的字符串来识别,并且可以在列表中的任何位置出现.
and remove or replace the /path/to/app/bin
without clobbering the rest of the variable. Extra points for allowing me put new elements in arbitrary positions. The target will be recognizable by a well defined string, and may occur at any point in the list.
我知道我已经看到了这一点,并且可以独自完成一些工作,但是我正在寻找一种不错的方法.可移植性和标准化是一个加分.
I know I've seen this done, and can probably cobble something together on my own, but I'm looking for a nice approach. Portability and standardization a plus.
我使用bash,但是在您喜欢的shell中也欢迎使用示例.
I use bash, but example are welcome in your favorite shell as well.
这里的上下文是一种需要在大型科学分析程序包的多个版本(一个用于进行分析,另一个用于框架)之间方便地切换的方法,该程序包可生成数十个可执行文件,数据保存在文件系统中,并且使用环境变量来帮助找到所有这些东西.我想编写一个选择版本的脚本,并且需要能够删除与当前活动版本有关的$PATH
元素,并用与新版本有关的相同元素替换它们.
The context here is one of needing to switch conveniently between multiple versions (one for doing analysis, another for working on the framework) of a large scientific analysis package which produces a couple dozen executables, has data stashed around the filesystem, and uses environment variable to help find all this stuff. I would like to write a script that selects a version, and need to be able to remove the $PATH
elements relating to the currently active version and replace them with the same elements relating to the new version.
这与在重新运行登录脚本等时防止重复$PATH
元素的问题有关.
This is related to the problem of preventing repeated $PATH
elements when re-running login scripts and the like.
- 先前的类似问题:如何避免复制路径变量在csh
- 随后的类似问题:
- Previous similar question: How to keep from duplicating path variable in csh
- Subsequent similar question: What is the most elegant way to remove a path from the $PATH variable in Bash?
推荐答案
从dmckee解决建议的解决方案:
Addressing the proposed solution from dmckee:
- 虽然某些版本的Bash可能允许在函数名称中使用连字符,但其他版本(MacOS X)则不允许.
- 我认为不需要在函数结束前立即使用return.
- 我看不到所有分号都需要.
- 我不明白为什么您要按路径逐个元素导出一个值.认为
export
等效于设置(甚至创建)全局变量-在任何可能的情况下都应避免. - 我不确定您希望'
replace-path PATH $PATH /usr
'做什么,但是它并没有达到我的期望.
- While some versions of Bash may allow hyphens in function names, others (MacOS X) do not.
- I don't see a need to use return immediately before the end of the function.
- I don't see the need for all the semi-colons.
- I don't see why you have path-element-by-pattern export a value. Think of
export
as equivalent to setting (or even creating) a global variable - something to be avoided whenever possible. - I'm not sure what you expect '
replace-path PATH $PATH /usr
' to do, but it does not do what I would expect.
考虑一个开始于以下内容的PATH值:
Consider a PATH value that starts off containing:
.
/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
我从"replace-path PATH $PATH /usr
"获得的结果是:
The result I got (from 'replace-path PATH $PATH /usr
') is:
.
/Users/jleffler/bin
/local/postgresql/bin
/local/mysql/bin
/Users/jleffler/perl/v5.10.0/bin
/local/bin
/bin
/bin
/sw/bin
/sbin
/sbin
由于/usr不会显示为(完整的)路径元素,而只是显示为路径元素的一部分,所以我本来希望恢复原来的路径.
I would have expected to get my original path back since /usr does not appear as a (complete) path element, only as part of a path element.
可以通过修改sed
命令之一来在replace-path
中修复该问题:
This can be fixed in replace-path
by modifying one of the sed
commands:
export $path=$(echo -n $list | tr ":" "\n" | sed "s:^$removestr\$:$replacestr:" |
tr "\n" ":" | sed "s|::|:|g")
我使用:"代替"|"分隔替换部分,因为"|"可以(理论上)出现在路径组件中,而根据PATH的定义,冒号不能出现.我观察到第二个sed
可以从PATH的中间消除当前目录.也就是说,PATH的合法值(尽管是反常的)可能是:
I used ':' instead of '|' to separate parts of the substitute since '|' could (in theory) appear in a path component, whereas by definition of PATH, a colon cannot. I observe that the second sed
could eliminate the current directory from the middle of a PATH. That is, a legitimate (though perverse) value of PATH could be:
PATH=/bin::/usr/local/bin
处理后,当前目录将不再位于PATH上.
After processing, the current directory would no longer be on the PATH.
在path-element-by-pattern
中进行锚定匹配的类似更改是合适的:
A similar change to anchor the match is appropriate in path-element-by-pattern
:
export $target=$(echo -n $list | tr ":" "\n" | grep -m 1 "^$pat\$")
我注意到grep -m 1
不是标准的(它是GNU扩展,也可在MacOS X上使用).而且,的确,echo
的-n
选项也是非标准的.您最好只删除通过将换行符从echo转换为冒号而添加的结尾冒号.由于按模式使用逐个路径的路径,具有不良的副作用(它掩盖了任何先前存在的称为$removestr
的导出变量),因此可以用其主体明智地替换它.加上更广泛地使用引号来避免空格或不必要的文件名扩展问题,会导致:
I note in passing that grep -m 1
is not standard (it is a GNU extension, also available on MacOS X). And, indeed, the-n
option for echo
is also non-standard; you would be better off simply deleting the trailing colon that is added by virtue of converting the newline from echo into a colon. Since path-element-by-pattern is used just once, has undesirable side-effects (it clobbers any pre-existing exported variable called $removestr
), it can be replaced sensibly by its body. This, along with more liberal use of quotes to avoid problems with spaces or unwanted file name expansion, leads to:
# 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 <grep pattern for target path>
#
# To replace a path:
# replace_path PATH $PATH /exact/path/to/remove /replacement/path
# replace_path_pattern PATH $PATH <target pattern> /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"
}
我有一个名为echopath
的Perl脚本,在调试类似PATH的变量的问题时,我发现它很有用:
I have a Perl script called echopath
which I find useful when debugging problems with PATH-like variables:
#!/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:
When I run the modified solution on the test code below:
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 $xpath "/usr/.*/bin" /work/bin
echopath xpath
输出为:
.
/Users/jleffler/bin
/usr/local/postgresql/bin
/usr/local/mysql/bin
/Users/jleffler/perl/v5.10.0/bin
/usr/local/bin
/usr/bin
/bin
/sw/bin
/usr/sbin
/sbin
.
/Users/jleffler/bin
/usr/local/postgresql/bin
/usr/local/mysql/bin
/Users/jleffler/perl/v5.10.0/bin
/usr/local/bin
/work/bin
/bin
/sw/bin
/usr/sbin
/sbin
.
/Users/jleffler/bin
/work/bin
/usr/local/mysql/bin
/Users/jleffler/perl/v5.10.0/bin
/usr/local/bin
/usr/bin
/bin
/sw/bin
/usr/sbin
/sbin
这对我来说似乎是正确的-至少对于我对问题的定义而言.
This looks correct to me - at least, for my definition of what the problem is.
我注意到echopath LD_LIBRARY_PATH
评估$LD_LIBRARY_PATH
.如果您的函数能够做到这一点,那就太好了,因此用户可以输入:
I note that echopath LD_LIBRARY_PATH
evaluates $LD_LIBRARY_PATH
. It would be nice if your functions were able to do that, so the user could type:
replace_path PATH /usr/bin /work/bin
可以使用以下方法完成:
That can be done by using:
list=$(eval echo '$'$path)
这导致对代码的修订:
# 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 <grep pattern for target path>
#
# To replace a path:
# replace_path PATH /exact/path/to/remove /replacement/path
# replace_path_pattern PATH <target pattern> /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" "$replacestr"
}
以下修订的测试现在也可以使用:
The following revised test now works too:
echo
xpath=$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.
It produces the same output as before.
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