checker284
checker284

Reputation: 1326

How-To get root directory of given path in bash?

My script:

    #!/usr/bin/env bash
    PATH=/home/user/example/foo/bar
    mkdir -p /tmp/backup$PATH

And now I want to get first folder of "$PATH": /home/

    cd /tmp/backup
    rm -rf ./home/
    cd - > /dev/null

How can I always detect the first folder like the example above? "dirname $PATH" just returns "/home/user/example/foo/".

Thanks in advance! :)

Upvotes: 24

Views: 84124

Answers (7)

Alcamtar
Alcamtar

Reputation: 1582

To get the first directory component of VAR:

echo ${VAR%${VAR#/*/}}

So, if VAR="/path/to/foo", this returns /path/.

Explanation:

${VAR#X} strips off the prefix X and returns the remainder. So if VAR=/path/to/foo, then /*/ matches the prefix /path/ and the expression returns the suffix to/foo.

${VAR%X} strips off the suffix X. By inserting the output of ${VAR#X}, it strips off the suffix and returns the prefix.

If you can guarantee that your paths are well formed this is a convenient method. It won't work well for some paths, such as //path/to/foo or path/to/foo, but you can handle such cases by breaking down the strings further.

Upvotes: 4

RobLoach
RobLoach

Reputation: 2206

You can use dirname...

#/usr/bin/env bash
DIRECTORY="/home/user/example/foo/bar"
BASE_DIRECTORY=$(dirname "${DIRECTORY}")
echo "#$BASE_DIRECTORY#";

Outputs the following...

/home/user/example/foo

Upvotes: -1

Thom
Thom

Reputation: 551

Pure bash:

DIR="/home/user/example/foo/bar"
[[ "$DIR" =~ ^[/][^/]+ ]] && printf "$BASH_REMATCH"

Easy to tweak the regex.

Upvotes: 0

condorwasabi
condorwasabi

Reputation: 626

You can try this awk command:

 basedirectory=$(echo "$PATH" | awk -F "/" '{print $2}')

At this point basedirectory will be the string home Then you write:

rm -rf ./"$basedirectory"/

Upvotes: 8

checker284
checker284

Reputation: 1326

I've found a solution:

    #/usr/bin/env bash
    DIRECTORY="/home/user/example/foo/bar"
    BASE_DIRECTORY=$(echo "$DIRECTORY" | cut -d "/" -f2)
    echo "#$BASE_DIRECTORY#";

This returns always the first directory. In this example it would return following:

    #home#

Thanks to @condorwasabi for his idea with awk! :)

Upvotes: 45

anubhava
anubhava

Reputation: 786329

To get the first firectory:

path=/home/user/example/foo/bar
mkdir -p "/tmp/backup$path"
cd /tmp/backup
arr=( */ )
echo "${arr[0]}"

PS: Never use PATH variable in your script as it will overrider default PATH and you script won't be able to execute many system utilities

EDIT: Probably this should work for you:

IFS=/ && set -- $path; echo "$2"
home

Upvotes: 1

konsolebox
konsolebox

Reputation: 75618

If PATH always has an absolute form you can do tricks like

ROOT=${PATH#/} ROOT=/${ROOT%%/*}

Or

IFS=/ read -ra T <<< "$PATH"
ROOT=/${T[1]}

However I should also add to that that it's better to use other variables and not to use PATH as it would alter your search directories for binary files, unless you really intend to.

Also you can opt to convert your path to absolute form through readlink -f or readlink -m:

ABS=$(readlink -m "$PATH")

You can also refer to my function getabspath.

Upvotes: 7

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