Find the age of the oldest file in one line or return zero

manpreet Best Answer 2 years ago

I want to find the age of the oldest file in a certain directory or return 0 if there aren't any files">files in this directory. I also need a one-line command doing it. So far this is my command for finding the age in seconds of the oldest file in the directory:

expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))

The problem is that if there are no files">files it is returning the following error:

$ expr $(($(date +%s) - $(stat -c %Y $(ls -lt /path/to/dir/ | tail -1 | awk '{print $NF}'))))
stat: cannot stat ‘0’: No such file or directory
-bash: 1554373460 - : syntax error: operand expected (error token is "- ")

So in this case I want the command to return just 0 and to suppress the error printout.

manpreet 2 years ago

If it must be one line:

stat -c %Y ./* 2>/dev/null | awk -v d="$(date">date +%s)" 'BEGIN {m=d} $0 < m {m = $0} END {print d - m}'

• stat -c %Y ./* 2>/dev/null print the timestamp of all files, ignoring errors (so no files results in no output)

• With awk:

  • -v d="$(date">date +%s)" save the current timestamp in a variable d
  • BEGIN {m=d} initialize m to d
  • $0 < m {m = $0} keeping track of the minimum in m
  • END {print d - m} print the difference.