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linuxbashpathls

Linux Bash Script: How to get file without path?

I am trying to write a very very simple script in Linux.
Let me show you the code first:

#!/bin/bash
# The shell program uses glob constructs and ls
# to list all entries in testfiles, that have 2
# or more dots "." in their name.

ls -l /path/to/file/*.*.*

When I run this code with bash myscript command, I get something like: /path/to/file/file.with.three.dots

But I don't want this. I want to show only the file name, not the path.
Then I tried: 

ls -l *.*.*

But this time is shows me the files, only if I am inside the /path/to/file/.
How can I set the path, so when running the script from any place, it will output the name of the files in the /path/to/file/?

Thank you!

Solution

  • basename path/to/file.b.c should give you file.b.c

    However re-reading the question, I think a temporary cd to the path and then an ls may be better:

    (cd /path/to/file; ls -l *.*.*)