使用 bash 从给定路径中删除不必要的斜线
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Remove unnecessary slashes from a given path with bash
提问by casper
How can I get rid of unnecessary slashes in a given path?
如何摆脱给定路径中不必要的斜线？
Example:
例子：
p="/foo//////bar///hello/////world"
I want:
我想要：
p="/foo/bar/hello/world"
回答by phihag
回答by Paused until further notice.
Using pure Bash:
使用纯 Bash：
shopt -s extglob
echo ${p//\/*(\/)/\/}
回答by Mattia72
With realpath:
使用真实路径：
realpath -sm $p
realpath -sm $p
Parameters:
参数：
-m, --canonicalize-missing no components of the path need exist
-s, --strip, --no-symlinks don't expand symlinks
回答by dsaydon
your input:
您的输入：
p="/foo//////bar///hello/////world"
command to remove the irrelevant slashes:
删除不相关斜线的命令：
echo $p | tr -s /
output:
输出：
/foo/bar/hello/world
回答by Feri
This works with multiple separators and does not assume the given path should exist:
这适用于多个分隔符，并且不假设给定的路径应该存在：
p=/foo///.//bar///foo1/bar1//foo2/./bar2;
echo $p | awk '{while(index(,"/./")) gsub("/./","/"); while(index(,"//"))
gsub("//","/"); print ;}'
But does not simplify well strings containing ".."
但并没有简化包含“..”的字符串
回答by Robie Basak
- Consider if you need to do this. On Unix, specifying duplicate path separators (and even things like
/foo/.//bar///hello/./worldwork just fine. - You can use
readlink -f, but this will also canonicalize the symlinks in that path, so the result depends on your filesystem and the path supplied must actually exist, so this won't work for virtual paths.
- 考虑您是否需要这样做。在 Unix 上，指定重复的路径分隔符（甚至像这样的东西
/foo/.//bar///hello/./world也很好用。 - 您可以使用
readlink -f，但这也将使该路径中的符号链接规范化，因此结果取决于您的文件系统，并且提供的路径必须实际存在，因此这不适用于虚拟路径。
回答by casper
Thanks for the replys. I know the path works fine. I just want this for optical reasons.
感谢您的回复。我知道路径工作正常。我只是出于光学原因想要这个。
I found another solution: echo $p | replace '//' ''
我找到了另一个解决方案： echo $p | replace '//' ''
