Shell Script to extract only file name

Question

I have file format like this (publishfile.txt)

drwxrwx---+  h655201 supergroup          0  2019-04-24  09:16  /data/xyz/invisible/se/raw_data/OMEGA
drwxrwx---+  h655201 supergroup          0  2019-04-24  09:16  /data/xyz/invisible/se/raw_data/sample
drwxrwx---+  h655201 supergroup          0  2019-04-24  09:16  /data/xyz/invisible/se/raw_data/sample(1)

I just want to extract the name OMEGA****, sample, sample(1) How can I do that I have used basename in my code but it doesn't work in for loop. Here is my sample code

for line in $(cat $BASE_PATH/publishfile.txt)
do 
      FILE_PATH=$(echo "line"| awk '{print  $NF}' )
done
FILE_NAME=($basename  $FILEPATH)

But this code also doesn't wor when used outside for loop

score 4 · Accepted Answer · answered Feb 02 '20 at 20:00

4

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

This simply says that the delimiter is a slash and we want the last field from each line.

You should basically never run Awk on each input line in a shell while read loop (let alone a for loop); Awk itself does this by default, much faster and better than the shell.

answered Feb 02 '20 at 20:00

tripleee

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this doesn't work if the file contains something like: `drwxrwx---+ h655201 supergroup 0 2019-04-24 09:16 sample` (i.e. a filepath without slash `/` char). You are correct with all other remarks! `awk` itself does the loop, etc. Using `awk` or `sed` is indeed a better solution! – azbarcea Feb 02 '20 at 20:06
Not directly, though it would not be very hard to adapt (try -F `[/ \t]'`) – tripleee Feb 02 '20 at 20:08
How can I use in a way that file name is separated by comma ? Like sample,sample1 – Shivam Sharma Feb 03 '20 at 10:25
1

See the section _Parameter Expansion_ in the [POSIX shell specification](https://pubs.opengroup.org/onlinepubs/009695399/utilities/xcu_chap02.html). If this doesn't answer your case, please ask a **new** question here, where you specify exactly the format of your input, and what you want to achieve. – user1934428 Feb 03 '20 at 11:00
If you are asking how to provide that sort o| output, the Awk script can easily be changed to use `printf` instead of `print`. If you are asking about input, I absolutely agree with the previous comment. – tripleee Feb 03 '20 at 11:11

score 1 · Answer 2 · answered Feb 02 '20 at 19:58

1

In your code above you have a typo. Your code reads:

    FILE_NAME=($basename  $FILEPATH)

but it should read

    FILE_NAME=$(basename  $FILEPATH)

That should work fine in or outside of a loop

answered Feb 02 '20 at 19:58

Rob Sweet

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8

I have tried FILE_NAME=$(basename $FILEPATH) this but it doesn't work – Shivam Sharma Feb 02 '20 at 20:00
Can you post the actual code where you tried that and it didn't work? – Rob Sweet Feb 02 '20 at 20:07
Properly speaking you should also quote `"$FILEPATH"` but of course if you kept the buggy `for` loop it will be too late to actually fix any quoting problems. – tripleee Feb 04 '20 at 04:29

azbarcea · Answer 3 · 2020-02-02T19:59:49.623

0

Try this:

cat $BASE_PATH/publishfile.txt | awk '{print $7}' | sed 's/.*\///'

the output will be:

OMEGA
sample
sample(1)

UPDATE: I guess cat x.txt | sed 's/.*\///' will still work, if all your files, folders contain at least 1 slash (/).

For the commands used, the manuals are: cat, awk, sed

edited Feb 02 '20 at 19:59

answered Feb 02 '20 at 19:55

azbarcea

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in the for loop line ? – Shivam Sharma Feb 02 '20 at 19:57
The [`cat` is useless.](https://stackoverflow.com/questions/11710552/useless-use-of-cat) – tripleee Feb 02 '20 at 20:09
yes, you can redirect the output, that's true, it may also be a little bit more efficient too. Probably depends how big the file is. – azbarcea Feb 02 '20 at 20:16

Shell Script to extract only file name

3 Answers3