巴什去除路径变量和sed命令路径 [英] Bash removing path from path variable and sed command

问题描述

本$ C $工作c

<一个href=\"http://stackoverflow.com/questions/370047/what-is-the-most-elegant-way-to-remove-a-path-from-the-path-variable-in-bash\">What是最优雅的方式来删除从猛砸$ PATH变量的路径？

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

在这种情况下，如果我只运行：

In this instance if I run just the:

export PATH=`echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}'`

我可以得到含 / SDE / 删除路径;然而，一个：仍然存在。我假设后的的sed 命令应该删除。当我一次运行没有东西都去掉这整个命令。有什么不对所导致的路径不更新，我怎样才能让 SED 命令删除 SED 语句结肠：在 / SDE / 路径变量被删除。

I can get the path containing /SDE/ removed; however a : remains. The sed command after I am assuming should remove that. When I run this entire command at once nothing gets removed at all. What is wrong with the sed statement that is causing the path not to update and how can I make the sed command remove the colon : after the /SDE/ path variable is removed.

推荐答案

的问题是关闭反引用的位置`中的命令：

The problem is the placement of the closing back-quote ` in the command:

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

如果您使用了推荐的 $（...）符号，你会看到，这相当于：

If you used the recommended $(...) notation, you'd see that this is equivalent to:

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

该管道导出操作的输出，以 SED ，但导出沉默。

which pipes the output of the export operation to sed, but export is silent.

使用：

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

我有固定的回答从错误的命令被逐字复制。正如评论 tripleee 的笔记，我不是完全由 AWK 的解决方案，但问题是什么是错的code'答案是'，其中反引号都放在'。在 AWK 脚本不会处​​理在路径中的任何位置删除路径的元素;在 SED 脚本只是确保没有尾随：，以便有结尾没有隐式使用当前目录的路径。

I have fixed the answer from which the erroneous command was copied verbatim. As tripleee notes in a comment, I'm not wholly convinced by the awk solution, but the question was 'what was wrong with the code' and the answer is 'where the back-quotes are placed'. The awk script does handle removing elements of a PATH at any position in the PATH; the sed script simply ensures there is no trailing : so that there is no implicit use of the current directory at the end of the PATH.

参见：<一href=\"http://stackoverflow.com/questions/273909/how-do-i-manipulate-path-elements-in-shell-scripts\">How我操纵shell脚本路径元素并在<一的 clnpath 脚本href=\"http://stackoverflow.com/questions/135754/how-to-keep-from-duplicating-path-variable-in-csh\">How从在 CSH 复制PATH变量保持 - 剧本是POSIX杂交炮弹像伯恩，科恩，击炮弹，尽管问题的主题。这里使用的标记 clnpath 之间的一个差异，并且是 clnpath 只删除完整路径;它不会尝试做局部路径元件匹配：

See also: How do I manipulate PATH elements in shell scripts and the clnpath script at How to keep from duplicating PATH variable in csh — the script is for POSIX-ish shells like the Bourne, Korn, Bash shells, despite the question's subject. One difference between clnpath and the notation used here is that clnpath only removes full pathnames; it does not attempt to do partial path element matching:

export PATH=$(clnpath $PATH /opt/SDE/bin)

如果要删除的路径元素是的/ opt / SDE /斌。需要注意的是 clnpath 可用于维持 LD_LIBRARY_PATH ， CDPATH ， MANPATH 和任何其他路径样变;所以可以在 AWK 调用，当然了。

if the path element to be removed was /opt/SDE/bin. Note that clnpath can be used to maintain LD_LIBRARY_PATH, CDPATH, MANPATH and any other path-like variable; so can the awk invocation, of course.

我顺便指出，在 AWK 脚本 / SDE / 模式将删除的/ opt / USDER /斌;在正则表达式的斜线都无关的路径斜杠。

I note in passing that that the /SDE/ pattern in the awk script will remove /opt/USDER/bin; the slashes in the regex have nothing to do with slashes in the pathname.

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