Bash从路径变量和sed命令中删除路径 [英] Bash removing path from path variable and sed command

问题描述

使用此代码

从Bash中的$ PATH变量中删除路径的最优雅的方式是什么？

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

在这种情况下，如果我运行： p>

In this instance if I run just the:

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

我可以得到包含 / SDE / 的路径;然而，：仍然存在。假设我应该删除 sed 命令。当我立即运行这个整个命令什么都没有被删除。导致路径不更新的 sed 语句有什么问题，如何使 sed 命令删除 / SDE / 路径变量删除后，冒号：

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

问题是关闭回报的位置`在命令中：

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

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

如果您使用推荐的 $（...）符号，你会发现这相当于：

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

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

哪些管道输出的操作到 sed ，但导出是静默的。

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

使用：

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

我已经修复了错误命令逐字复制的答案。作为评论中的三重符号注释，我并不完全相信 awk 解决方案，但问题是代码有什么问题，答案是后置引号放在哪里。 awk 脚本确实在PATH中的任何位置处理删除PATH的元素; sed 脚本只是确保没有尾随：，以便最终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.

另请参见：如何在shell脚本中操作PATH元素和 clnpath 脚本在如何避免在中重复PATH变量csh - 脚本是用于POSIX-ish贝壳像Bourne，Korn，Bash shell，尽管有问题的主题。 clnpath 与此处使用的符号之间的区别是 clnpath 仅删除完整的路径名;它不会尝试执行部分路径元素匹配：

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为 / opt / SDE / bin 。请注意， clnpath 可用于维护 LD_LIBRARY_PATH ， CDPATH ， MANPATH 和任何其他类似路径的变量;所以当然可以 awk 调用。

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.

我注意到， / code $ c>脚本中的/ SDE / 模式将删除 / opt / USDER / bin ;正则表达式中的斜杠与路径名中的斜杠无关。

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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