Get last field using awk substr

Question

I am trying to use awk to get the name of a file given the absolute path to the file.
For example, when given the input path /home/parent/child/filena

Answer 1

If you're open to a Perl solution, here one similar to fedorqui's awk solution:

perl -F/ -lane 'print $F[-1]' input

-F/ specifies / as the field separator
$F[-1] is the last element in the @F autosplit array

Answer 2

It should be a comment to the basename answer but I haven't enough point.

If you do not use double quotes, basename will not work with path where there is space character:

$ basename /home/foo/bar foo/bar.png
bar

ok with quotes " "

$ basename "/home/foo/bar foo/bar.png"
bar.png

file example

$ cat a
/home/parent/child 1/child 2/child 3/filename1
/home/parent/child 1/child2/filename2
/home/parent/child1/filename3

$ while read b ; do basename "$b" ; done < a
filename1
filename2
filename3

Answer 3

In this case it is better to use basename instead of awk:

 $ basename /home/parent/child1/child2/filename
 filename

Answer 4

I know I'm like 3 years late on this but.... you should consider parameter expansion, it's built-in and faster.

if your input is in a var, let's say, $var1, just do ${var1##*/}. Look below

$ var1='/home/parent/child1/filename'
$ echo ${var1##*/}
filename
$ var1='/home/parent/child1/child2/filename'
$ echo ${var1##*/}
filename
$ var1='/home/parent/child1/child2/child3/filename'
$ echo ${var1##*/}
filename

Answer 5

You can also use:

    sed -n 's/.*\/\([^\/]\{1,\}\)$/\1/p'

or

    sed -n 's/.*\/\([^\/]*\)$/\1/p'

Answer 6

Like 5 years late, I know, thanks for all the proposals, I used to do this the following way:

$ echo /home/parent/child1/child2/filename | rev | cut -d '/' -f1 | rev
filename

Glad to notice there are better manners