Remove superfluous slashes from a given path with bash

Question

How can I get rid of unnecessary slashes in a given path?

Example:

p="/foo//////bar///hello/////world" 

I want:

 p="/foo/bar/hello/world" 

Answer 1

Use readlink :

 p=$(readlink -m "/foo//////bar///hello/////world") 

Please note that this will lead to the canonization of symbolic links. If this is not what you want, use sed :

 p=$(echo "/foo//////bar///hello/////world" | sed s#//*#/#g) 

Answer 2

Using pure bash:

 shopt -s extglob echo ${p//\/*(\/)/\/} 

Answer 3

With realpath:

realpath -sm $p

Options:

  -m, --canonicalize-missing no components of the path need exist -s, --strip, --no-symlinks don't expand symlinks 

Answer 4

• Think about whether you need to do this. On Unix, specifying duplicate path separators (and even things like /foo/.//bar///hello/./world work fine.
• You can use readlink -f , but this will also lead to canonicalization of symbolic links in this path, so the result will depend on your file system, and the provided path must actually exist, so this will not work for virtual paths.

Answer 5

This works with multiple delimiters and does not assume that this path should exist:

 p=/foo///.//bar///foo1/bar1//foo2/./bar2; echo $p | awk '{while(index($1,"/./")) gsub("/./","/"); while(index($1,"//")) gsub("//","/"); print $1;}' 

But does not simplify lines containing ".."

Answer 6

your input:

 p="/foo//////bar///hello/////world" 

to remove irrelevant slashes:

 echo $p | tr -s / 

exit:

 /foo/bar/hello/world 

Answer 7

Thanks for answers. I know the way works fine. I just want this for optical reasons.

I found another solution: echo $p | replace '//' '' echo $p | replace '//' ''