Get just the filename from a path in a Bash script [duplicate]

Many UNIX-like operating systems have a basename executable for a very similar purpose (and dirname for the path):

pax> full_name=/tmp/file.txt
pax> base_name=$(basename ${full_name})
pax> echo ${base_name}
file.txt

That unfortunately just gives you the file name, including the extension, so you’d need to find a way to strip that off as well.

So, given you have to do that anyway, you may as well find a method that can strip off the path and the extension.

One way to do that (and this is a bash-only solution, needing no other executables):

pax> full_name=/tmp/xx/file.tar.gz
pax> xpath=${full_name%/*} 
pax> xbase=${full_name##*/}
pax> xfext=${xbase##*.}
pax> xpref=${xbase%.*}
pax> echo "path="${xpath}", pref="${xpref}", ext="${xfext}""

path="/tmp/xx", pref="file.tar", ext="gz"

That little snippet sets xpath (the file path), xpref (the file prefix, what you were specifically asking for) and xfext (the file extension).

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