Remove part of path on Unix

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-
# output:
# some/deepish/dir

To keep the first two components:

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

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

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

To keep the last three components:

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

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

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

Or after a particular component:

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

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

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

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

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

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

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

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

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.