ksh storing result of a command to a variable

Question

I want to store the result of a command to a variable in my shell script. I cant seem to get it to work. I want the most recently dated file in the directory.

Answer 1

The problem that you're having is that the command needs to be surrounded by back-ticks rather than single quotes. This is known as 'Command Substitution'.

Bash allows you to use $() for command substitution, but this is not available in all shells. I don't know if it's available in KSH; if it is, it's probably not available in all versions.

If the $() syntax is available in your version of ksh, you should definitely use it; it's easier to read (back ticks are too easy to confuse with single quotes); back-ticks are also hard to nest.

This only addresses one of the problems with your command, however: ls returns directories as well as files, so if the most recent thing modified in the specified directory is a sub-directory, that is what you will see.

If you only want to see files, I suggest using some version of the following (I'm using Bash, which supports default variables, you'll probably have to play around with the syntax of $1)

lastfile () 
{ 
    find ${1:-.} -maxdepth 1 -type f -printf "%T+ %p\n" | sort -n | tail -1 | sed 's/[^[:space:]]\+ //'
}

This runs find on the directory, and only pulls files from that directory. It formats all of the files like this:

2012-08-29+16:21:40.0000000000 ./.sqlite_history
2013-01-14+08:52:14.0000000000 ./.davmail.properties
2012-04-04+16:16:40.0000000000 ./.DS_Store
2010-04-21+15:49:00.0000000000 ./.joe_state
2008-09-05+17:15:28.0000000000 ./.hplip.conf
2012-01-31+13:12:28.0000000000 ./.oneclick

sorts the list, takes the last line, and chops off everything before the first space.