To show only file name without the entire directory path

ls /home/user/new/*.txt prints all txt files in that directory. However it prints the output as follows:

[me@comp]$ ls /home/user/new/*.txt
/home/user/new/file1.txt    /home/user/new/file2.txt    /home/user/new/file3.txt

and so on.

I want to run the ls command not from the /home/user/new/ directory thus I have to give the full directory name, yet I want the output to be only as

[me@comp]$ ls /home/user/new/*.txt
file1.txt    file2.txt    file3.txt 

I don't want the entire path. Only filename is needed. This issues has to be solved using ls command, as its output is meant for another program.

Solution 1:

ls whateveryouwant | xargs -n 1 basename

Does that work for you?

Otherwise you can (cd /the/directory && ls) (yes, parentheses intended)

Solution 2:

No need for Xargs and all , ls is more than enough.

ls -1 *.txt

displays row wise

Solution 3:

There are several ways you can achieve this. One would be something like:

for filepath in /path/to/dir/*
do
    filename=$(basename $filepath)

    ... whatever you want to do with the file here
done

Solution 4:

Use the basename command:

basename /home/user/new/*.txt

Solution 5:

(cd dir && ls)

will only output filenames in dir. Use ls -1 if you want one per line.

(Changed ; to && as per Sactiw's comment).