how to print tail of path filename using awk

Question

I've searched it with no success.

I have a file with pathes. I want to print the tail of a all pathes. for example (for every line in file):

/homes/work/abc.txt
--> abc.txt

Does anyone know how to do it?

Thanks

What would you want to happen if your file contained `/usr/bin/`? — Ed Morton, Nov 11 '13 at 16:26

score 7 · Accepted Answer · answered Nov 10 '13 at 14:34

7

awk -F "/" '{print $NF}' input.txt

will give output of:

abc1.txt
abc2.txt
abc3.txt

for:

$>cat input.txt
text path/to/file/abc1.txt
path/to/file/abc2.txt
path/to/file/abc3.txt

answered Nov 10 '13 at 14:34

1

Can be shorten some `awk -F/ '{print $NF}'` – Jotne Nov 10 '13 at 16:44
3

@Jotne: or `awk -F/ '$0=$NF'` – Thor Nov 10 '13 at 18:46
@Thor Very nice indeed – Jotne Nov 10 '13 at 19:58
I second that! Nice to know – EverythingRightPlace Nov 10 '13 at 20:00
1

Thanks! what $NF stand for? – Noam Mizrachi Nov 11 '13 at 07:19
this is an old post... but I was wondering, too... [what does $NF stand for?](http://www.grymoire.com/Unix/Awk.html#uh-17) apparently, it is the number fields variable. – gloomy.penguin Oct 25 '14 at 17:10
1

@Thor This `awk -F/ '$0=$NF'` will fail for value `0`, so it need to be `awk -F/ '{$0=$NF}1'` (see my updated post) – Jotne Oct 25 '14 at 18:54
@gloomy.penguin: `NF` is the number of fields, so `$NF` refers to the last field. @jotne: This is true for this rare case and indeed the safer way to go. – Thor Oct 26 '14 at 03:43

Jotne · Answer 2 · 2014-10-25T18:52:37.850

5

How about this awk

echo "/homes/work/abc.txt" | awk '{sub(/.*\//,x)}1'
abc.txt

Since .* is greedy, it will continue until last /
So here we remove all until last / with x, and since x is empty, gives nothing.

Thors version

echo "/homes/work/abc.txt" | awk -F/ '$0=$NF'
abc.txt

NB this will fail for /homes/work/0 or 0,0 etc so better use:

echo "/homes/work/abc.txt" | awk -F/ '{$0=$NF}1'

edited Oct 25 '14 at 18:52

answered Nov 10 '13 at 14:23

Jotne

jkshah · Answer 3 · 2013-11-11T16:28:55.107

4

awk solutions are already provided by @Jotne and @bashophil

Here are some other variations (just for fun)

Using sed

sed 's:.*/::' file

Using grep

grep -oP '(.*/)?\K.*' file

Using cut - added by @Thor

rev file | cut -d/ -f1 | rev

Using basename - suggested by @fedorqui and @EdMorton

while IFS= read -r line; do 
  basename "$line" 
done < file

edited Nov 11 '13 at 16:28

answered Nov 10 '13 at 14:28

jkshah

@fedorqui Thanks for your suggestion! Added to solution. – jkshah Nov 11 '13 at 10:59
Note in my proposed version it won't fail if file paths have white spaces, because quoting the variable when calling `basename` avoids it. – fedorqui Nov 11 '13 at 11:01
@fedorqui that's cool. actually i had tried the same earlier but might have missed quotes so decided to go with `xargs` as produced results were same. Now I have removed that bad practice. This is *fun*. Thanks :) – jkshah Nov 11 '13 at 11:07
The loop will fail if your file names contain backslashes or can start with white space. Always us `while IFS= read -r line` unless you have a very specific reason not to. Obviously, the approach just won't work at all if the file names contain newlines but that's a whole other can of worms. – Ed Morton Nov 11 '13 at 16:25

3 Answers3