Quote:

Originally Posted by otheus Actually, Gee-Money is on the ball. If your script is in a particular directory, you don't need find. Just ls. However, his script is unnecessarily complex: Code: ls -1t |sed -n '/name.*\.log$/{p;q}' would do and be a bit quicker. (Normally, ls prints out in columns UNLESS it's printing out to a pipe. The -1 here just emphasizes that.)

No, ls only prints in columns if the output is going to a terminal. Anywhere else (e.g., pipe or redirection), it prints one file per line.

Quote:

Originally Posted by gugs Our system produce logs when a script is run which may not be daily, the logs have a format: name_YYMMDD.log - both name and .log are consistent, date changes as per the day the script is run. Is there a way of finding the last saved log?

Since the shell will sort the files automatically, there is no need for any external command:

Code:

set -- name_*.log
shift $(( $# - 1 ))
lastlog=$1

There's nothing cheap about using three external commands when you don't need any.

Quote:

Originally Posted by treesloth Considering the extremely lightweight task being performed, even three external commands are, in fact, extremely cheap with system resources.

Three external commands are always much more expensive in terms of time and systems resources than none.

External commands use many times more resources and time than shell builtin functions and syntax.

If the tasks are lightweight, that's all the more reason for not using heavy external commands.