How do I run a program only once per day, while accounting for variable uptime and possible failure of program?

Question

I need to run a command once, but only once, per day, until it succeeds.

Continuous uptime cannot be expected, and program success cannot be guaranteed.

This program requires network access, but not every time I start my computer with network access.

My program will exit with, e.g -1 unless it succeed ( which returns 0 ).

Answer 1

Use a shell to provide this. For example, create a script with something like the following:

#!/bin/sh
# Check to see if this is already running from some other day
mkdir /tmp/lock || exit 1
while ! command-to-execute-until-succeed; do
    # Wait 30 seconds between successive runs of the command
    sleep 30
done
rmdir /tmp/lock

After that, point cron to the script.

Answer 2

I think the easiest way to do this would be for your program to check for a file's existence before making the attempt, and to create that file once it executes successfully.

If you can't modify the program to do so, use the cronjob itself. Something along the lines of:

test -e /path/to/tempfile || ( /path/to/program && touch -a /path/to/tempfile )

Because of the way unix evaluates condition checks, once the left side of the || condition evaluates to true, the entire expression is true and the right side is never evaluated -- IE, the program isn't called.

Then inside the parens, && requires both sides to evaluate to true, so if /path/to/program exits with a nonzero return code, there's no point in evaluating the statement on the right -- the expression already evaluates to false -- and /path/to/tempfile is never created.