How do I remove the file suffix and path portion from a path string in Bash?

Given a string file path such as /foo/fizzbuzz.bar, how would I use bash to extract just the fizzbuzz portion of said string?

Solution 1:

Here's how to do it with the # and % operators in Bash.

$ x="/foo/fizzbuzz.bar"
$ y=${x%.bar}
$ echo ${y##*/}
fizzbuzz

${x%.bar} could also be ${x%.*} to remove everything after a dot or ${x%%.*} to remove everything after the first dot.

Example:

$ x="/foo/fizzbuzz.bar.quux"
$ y=${x%.*}
$ echo $y
/foo/fizzbuzz.bar
$ y=${x%%.*}
$ echo $y
/foo/fizzbuzz

Documentation can be found in the Bash manual. Look for ${parameter%word} and ${parameter%%word} trailing portion matching section.

Solution 2:

look at the basename command:

NAME="$(basename /foo/fizzbuzz.bar .bar)"

instructs it to remove the suffix .bar, results in NAME=fizzbuzz

Solution 3:

Pure bash, done in two separate operations:

1. Remove the path from a path-string:

   path=/foo/bar/bim/baz/file.gif
   
   file=${path##*/}  
   #$file is now 'file.gif'
   

2. Remove the extension from a path-string:

   base=${file%.*}
   #${base} is now 'file'.
   

Solution 4:

Using basename I used the following to achieve this:

for file in *; do
    ext=${file##*.}
    fname=`basename $file $ext`

    # Do things with $fname
done;

This requires no a priori knowledge of the file extension and works even when you have a filename that has dots in it's filename (in front of it's extension); it does require the program basename though, but this is part of the GNU coreutils so it should ship with any distro.