Reputation: 10063
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 $5
which would be incorrect if my input has the following structure:
/home/parent/child1/child2/filename
So a generic solution requires always taking the last field (which will be the filename).
Is there a simple way to do this with the awk substr function?
Upvotes: 159
Views: 241813
Reputation: 2811
just realized it's much much cleaner this way in mawk/gawk/nawk
:
echo '/home/parent/child1/child2/filename' | …
'
awk ++NF FS='.+/' OFS= # updated such that
# root "/" still gets printed
'
filename
Upvotes: 0
Reputation: 2811
you can skip all of that complex regex :
echo '/home/parent/child1/child2/filename' |
mawk '$!_=$-_=$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`
Upvotes: 0
Reputation: 1405
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
Upvotes: 5
Reputation: 1699
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
Upvotes: 3
Reputation: 289525
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
Upvotes: 321
Reputation: 167
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
Upvotes: 3
Reputation: 3451
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
Upvotes: 7
Reputation: 49
You can also use:
sed -n 's/.*\/\([^\/]\{1,\}\)$/\1/p'
or
sed -n 's/.*\/\([^\/]*\)$/\1/p'
Upvotes: -1
Reputation: 123458
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
Upvotes: 5
Reputation: 26164
In this case it is better to use basename
instead of awk
:
$ basename /home/parent/child1/child2/filename
filename
Upvotes: 39