list command grep current date output filename only

Question

I would like to know how to list a file by current date and output filename only

my current command:

ls -l /var/www/html/test --time-style=+%D  | grep $(date +%D)

output:

-rwxrwxr-x 1 root test  146 04/03/18 file1.txt
-rwxrwxr-x 1 root test  146 04/03/18 file2.txt

I tried ls /var/www/html/test --time-style=+%D | grep $(date +%D), and it didn't work.

basically, I want the output as filename only file1.txt and file2.txt

Do you want to output a list of all files, sorted by date? Or just files with a specific date? — match, Apr 03 '18 at 20:24

score 1 · Accepted Answer · edited Apr 04 '18 at 06:05

1

ls -l /var/www/html/test --time-style=+%D | grep $(date +%D) | awk '{print $7}'

edit: formatting code

edited Apr 04 '18 at 06:05

kyodev

answered Apr 03 '18 at 20:22

soft87

score 0 · Answer 2 · answered Apr 03 '18 at 23:42

0

I believe find command is better for the task. Read this answer.

find /var/www/html/test -type f -newermt $(date +%F)

I think it is more elegant solution.

answered Apr 03 '18 at 23:42

Dudi Boy

2 Answers2