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

Answer 1

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

edit: formatting code

Answer 2

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.