Extract file basename without path and extension in bash [duplicate]

Given file names like these:

/the/path/foo.txt
bar.txt

I hope to get:

foo
bar

Why this doesn't work?

#!/bin/bash

fullfile=$1
fname=$(basename $fullfile)
fbname=${fname%.*}
echo $fbname

What's the right way to do it?

Solution 1:

You don't have to call the external basename command. Instead, you could use the following commands:

$ s=/the/path/foo.txt
$ echo "${s##*/}"
foo.txt
$ s=${s##*/}
$ echo "${s%.txt}"
foo
$ echo "${s%.*}"
foo

Note that this solution should work in all recent (post 2004) POSIX compliant shells, (e.g. bash, dash, ksh, etc.).

Source: Shell Command Language 2.6.2 Parameter Expansion

More on bash String Manipulations: http://tldp.org/LDP/LG/issue18/bash.html

Solution 2:

The basename command has two different invocations; in one, you specify just the path, in which case it gives you the last component, while in the other you also give a suffix that it will remove. So, you can simplify your example code by using the second invocation of basename. Also, be careful to correctly quote things:

fbname=$(basename "$1" .txt)
echo "$fbname"

Solution 3:

A combination of basename and cut works fine, even in case of double ending like .tar.gz:

fbname=$(basename "$fullfile" | cut -d. -f1)

Would be interesting if this solution needs less arithmetic power than Bash Parameter Expansion.