Displaying file name using special characters-CodePudding

I've one file ABC_123.csv in my app directory and I want to display its full name. I found two ways to do it (see below code snippet): one using ??? and the other using asterisk at the end of required text ABC_.

But, both ways are also displaying the path along with the name. Both below commands are producing results in this format: path name. I only need the name. Is there any special character (like ? or *) to display the name file only?

[input]$ ls /usr/opt/app/ABC_???.csv
[output] /usr/opt/app/ABC_123.csv

[input]$ ls /usr/opt/app/ABC_*.csv
[output] /usr/opt/app/ABC_123.csv

I cannot do this:

[input]$ cd /usr/opt/app
[input]$ ls ABC_???.csv
[output] ABC_123.csv

Required output:

[input]$ ls /usr/opt/app/ABC_(some-special-character).csv
[output] ABC_123.csv

[Edited] basename is working, but, I want to achieve this using ls and some special character (as highlighted above in Required output). Is there any way to this?

[input]$ basename /usr/opt/app/ABC_???.csv
[output] ABC_123.csv

CodePudding user response：

You can use

find /usr/opt/app/ -type f -name "ABC_*.csv" -exec basename '{}' \;

basename will isolate the file name. find will search the specified directory for files that match the provided pattern.

CodePudding user response：

If you were limited to ls, then this might help.

file="$(echo /usr/opt/app/ABC_???.csv)"; echo "${file##*/}"

CodePudding user response：

Pipe every csv file to basename:

ls /usr/opt/app/ABC_*.csv | xargs basename -a

CodePudding user response：

Without a need for basename:

(cd /usr/opt/app/ && ls ABC_*.csv)

"I cannot do this" was similar but didn't explain why, so maybe one-liner is doable. Doing in sub-shell prevents current dir from changing.

And no - there is no special character that could be used there. It's globbing: https://en.wikipedia.org/wiki/Glob_(programming)