Remove part of path on Unix

问题:

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 /path/to/file/drive and produce the output:

file/path/

Note: I have several paths in a while loop, with the same /path/to/file/drive in all of them, but I'm just looking for the 'how to' on removing the desired string.

I found some examples, but I can't get them to work:

echo /path/to/file/drive/file/path/ | sed 's:/path/to/file/drive:\2:'
echo /path/to/file/drive/file/path/ | sed 's:/path/to/file/drive:2'

\2 being the second part of the string and I'm clearly doing something wrong...maybe there is an easier way?

回答1:

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.

回答2:

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.

回答3:

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/

回答4:

One way to do this with sed is

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

回答5:

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.

回答6:

If you want to remove the first N parts of the path, you could of course use N calls to basename, 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".

回答7:

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