bash script: how to display output in different line

Question

i'm trying use one line of code to solve a problem

echo $(find . -maxdepth 1 -type f -newer $1  | sed 's,\.\/,,g')

this will print out all the file in current folder that are newer than the input file. But it prints out in one single line:

file1 file2 file3 file4....

how can i display each file name in a single line like:

file1
file2
file3
...

This seems to be a very simple but i've been searching and have no solution. Thank you in advance.

John Kugelman · Accepted Answer · 2015-01-22 19:05:29Z

6

Get rid of the echo and the $(...).

find . -maxdepth 1 -type f -newer "$1" | sed 's,\.\/,,g'

If you have GNU find you can replace the sed with a -printf action:

find . -maxdepth 1 -type f -newer "$1" -printf '%P\n'

answered Jan 22, 2015 at 19:05

John Kugelman

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Comments

Aman · Accepted Answer · 2015-01-22 19:13:57Z

0

pipe it to tr:

... | tr " " "\n"

answered Jan 22, 2015 at 19:13

Aman

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