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/filename I would like to get filename I have tried:
awk -F "/" '{print $5}' input which works perfectly.
However, I am hard coding $5 which would be incorrect if my input has the following structure:
/home/parent/child1/child2/filename So a generic solution requires always taking the last field (which will be the filename).
Is there a simple way to do this with the awk substr function?
19 Answers
Use the fact that awk splits the lines in fields based on a field separator, that you can define. Hence, defining the field separator to / you can say:
awk -F "/" '{print $NF}' input as NF refers to the number of fields of the current record, printing $NF means printing the last one.
So given a file like this:
/home/parent/child1/child2/child3/filename /home/parent/child1/child2/filename /home/parent/child1/filename This would be the output:
$ awk -F"/" '{print $NF}' file filename filename filename In this case it is better to use basename instead of awk:
$ basename /home/parent/child1/child2/filename filename 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
Another option is to use bash parameter substitution.
$ foo="/home/parent/child/filename" $ echo ${foo##*/} filename $ foo="/home/parent/child/child2/filename" $ echo ${foo##*/} filename 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 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 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
you can skip all of that complex regex :
echo '/home/parent/child1/child2/filename' | mawk '$!_=$-_=$NF' FS='[/]' filename 2nd to last :
mawk '$!--NF=$NF' FS='/' child2 3rd last field :
echo '/home/parent/child1/child2/filename' | mawk '$!--NF=$--NF' FS='[/]' child1 4th-last :
mawk '$!--NF=$(--NF-!-FS)' FS='/' echo '/home/parent/child000/child00/child0/child1/child2/filename' | child0 echo '/home/parent/child1/child2/filename' parent major caveat :
- `gawk/nawk` has a slight discrepancy with `mawk` regarding - how it tracks multiple, - and potentially conflicting, decrements to `NF`, - so other than the 1st solution regarding last field, - the rest for now, are only applicable to `mawk-1/2` You can also use:
sed -n 's/.*\/\([^\/]\{1,\}\)$/\1/p' or
sed -n 's/.*\/\([^\/]*\)$/\1/p' 
