Shell Programming and Scripting

I have a file that is created via a perl script where the file is named like so: 01-07-2016_10:17:08. I am running a shell script that needs to take this file and print it. I can capture the date portion fine, but I am unsure how to capture the time stamp, since there will be a difference from what time the file is created and when I run another command to capture the time. There will only be one file created for each day. I feel like I should use some sort of regex, but I don't want to force the shell script to look through the entire directory for that file. Is there a way to do this?

In any way the shell will have to look through the directory anyway, because file regexes ("globs" in shell speak) are expanded by the shell - always. If you write a command like ls -l foo* The shell will first expand foo* to a list of files matching this expression, let us say fooA fooB fooC, and then feed this to the ls command. Entering the command ls -l fooA fooB fooC would have had the same effect.

Therefore: either you know exactly what your file is named beforehand or you will have to use a glob (like "*", "?", etc.). On the up side it doesn't take much time to search through a directory at all, because UNIX kernels are in the habit of putting every ounce of spare memory to disk caching. Because metadata (inodes, directories, the supernode and so on) are used very often (compared to the content of most files) chances are the needed data are already in memory and are found pretty quickly.

I hope this helps.

bakunin