删除 Unix 上的部分路径

One way to do this with sed is

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

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.

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/

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.

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".

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.

Pure bash, without hard coding the answer

basenames()
{
local d="${2}"
for ((x=0; x<"${1}"; x++)); do
d="${d%/*}"
done
echo "${2#"${d}"/}"
}

• Argument 1 - How many levels do you want to keep (2 in the original question)
• Argument 2 - The full path

Taken from vsi_common(original version)

Here's a solution using simple bash syntax that accommodates variables (in case you don't want to hard code full paths), removes the need for piping stdin to sed, and includes a for loop, for good measure:

FULLPATH="/path/to/file/drive/file/path/"
SUBPATH="/path/to/file/drive/"
for i in $FULLPATH;
do
echo ${i#$SUBPATH}
done

as mentioned above by @evil otto, the # symbol is used to remove a prefix in this scenario.