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

How would I get just the filename without the extension and no path?

The following gives me no extension, but I still have the path attached:

source_file_filename_no_ext=${source_file%.*}

Solution 1:

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).

Solution 2:

basename and dirname solutions are more convenient. Those are alternative commands:

FILE_PATH="/opt/datastores/sda2/test.old.img"
echo "$FILE_PATH" | sed "s/.*\///"

This returns test.old.img like basename.

This is salt filename without extension:

echo "$FILE_PATH" | sed -r "s/.+\/(.+)\..+/\1/"

It returns test.old.

And following statement gives the full path like dirname command.

echo "$FILE_PATH" | sed -r "s/(.+)\/.+/\1/"

It returns /opt/datastores/sda2