Extract Directory from Path [Duplicate]
In My Script I Need the Directory of the File I Am Working With. for Example, the File="Stuff/Backup/File. Zip". I Need a Way to Get the String "Stuff/Backup/"...
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In my script I need the directory of the file I am working with. For example, the file="stuff/backup/file.zip". I need a way to get the string "stuff/backup/" from the variable $file.
5 Answers
dirname $file
is what you are looking for
dirname $file
will output
stuff/backup
which is the opposite of basename:
basename $file
would output
file.zip
Using ${file%/*} like suggested by Urvin/LuFFy is technically better since you won't rely on an external command. To get the basename in the same way you could do ${file##*/}. It's unnecessary to use an external command unless you need to.
file="/stuff/backup/file.zip"
filename=${1##*/} # file.zip
directory=${1%/*} # /stuff/backup
It would also be fully POSIX compliant this way. Hope it helps! :-)
For getting directorypath from the filepath:
file="stuff/backup/file.zip"
dirPath=${file%/*}/
echo ${dirPath}
Simply use $ dirname /home/~username/stuff/backup/file.zip
It will return /home/~username/stuff/backup/
Marcus Vance is a cybersecurity auditor and technology writer dedicated to educating the public about online safety, data privacy regulations, enterprise security, and emerging cyber threats.
