Remove part of path on Unix

Question

I\'m trying to remove part of the path in a string. I have the path:

/path/to/file/drive/file/path/

I want to remove the first part /

Answer 1

Using ${path#/path/to/file/drive/} as suggested by evil otto is certainly the typical/best way to do this, but since there are many sed suggestions it is worth pointing out that sed is overkill if you are working with a fixed string. You can also do:

echo $PATH | cut -b 21-

To discard the first 20 characters. Similarly, you can use ${PATH:20} in bash or $PATH[20,-1] in zsh.

Answer 2

If you don't want to hardcode the part you're removing:

$ s='/path/to/file/drive/file/path/'
$ echo ${s#$(dirname "$(dirname "$s")")/}
file/path/

Answer 3

You can also use POSIX shell variable expansion to do this.

path=/path/to/file/drive/file/path/
echo ${path#/path/to/file/drive/}

The #.. part strips off a leading matching string when the variable is expanded; this is especially useful if your strings are already in shell variables, like if you're using a for loop. You can strip matching strings (e.g., an extension) from the end of a variable also, using %.... See the bash man page for the gory details.

Answer 4

If you wanted to remove a certain NUMBER of path components, you should use cut with -d'/'. For example, if path=/home/dude/some/deepish/dir:

To remove the first two components:

# (Add 2 to the number of components to remove to get the value to pass to -f)
$ echo $path | cut -d'/' -f4-
some/deepish/dir

To keep the first two components:

$ echo $path | cut -d'/' -f-3
/home/dude

To remove the last two components (rev reverses the string):

$ echo $path | rev | cut -d'/' -f4- | rev
/home/dude/some

To keep the last three components:

$ echo $path | rev | cut -d'/' -f-3 | rev
some/deepish/dir

Or, if you want to remove everything before a particular component, sed would work:

$ echo $path | sed 's/.*\(some\)/\1/g'
some/deepish/dir

Or after a particular component:

$ echo $path | sed 's/\(dude\).*/\1/g'
/home/dude

It's even easier if you don't want to keep the component you're specifying:

$ echo $path | sed 's/some.*//g'
/home/dude/

And if you want to be consistent you can match the trailing slash too:

$ echo $path | sed 's/\/some.*//g'
/home/dude

Of course, if you're matching several slashes, you should switch the sed delimiter:

$ echo $path | sed 's!/some.*!!g'
/home/dude

Note that these examples all use absolute paths, you'll have to play around to make them work with relative paths.

Answer 5

One way to do this with sed is

echo /path/to/file/drive/file/path/ | sed 's:^/path/to/file/drive/::'

Answer 6

If you want to remove the first N parts of the path, you could of course use N calls to dirname, as in glenn's answer, but it's probably easier to use globbing:

path=/path/to/file/drive/file/path/
echo "${path#*/*/*/*/*/}"   #  file/path/

Specifically, ${path#*/*/*/*/*/} means "return $path minus the shortest prefix that contains 5 slashes".