How to run a script at a certain time on Linux? [c

问题:

I have a text file containing a specific date and time. I want to be able to run a script at the time specified in that file. How would you achieve that? Create another script that runs in background (sort of a deamon) and checks every second if the current time is matching the time in the file? Is there another way? The machine is a linux server , Debian wheezy. Thanks in advance

回答1:

Look at the following:

echo "ls -l" | at 07:00

This code line executes "ls -l" at a specific time. This is an example of executing something (a command in my example) at a specific time. "at" is the command you were really looking for. You can read the specifications here:

http://manpages.ubuntu.com/manpages/precise/en/man1/at.1posix.html http://manpages.ubuntu.com/manpages/xenial/man1/at.1posix.html

Hope it helps!

回答2:

The at command exists specifically for this purpose (unlike cron which is intended for scheduling recurring tasks).

at $(cat file) </path/to/script

回答3:

Cron is good for something that will run periodically, like every Saturday at 4am. There's also anacron, which works around power shutdowns, sleeps, and whatnot. As well as at.

But for a one-off solution, that doesn't require root or anything, you can just use date to compute the seconds-since-epoch of the target time as well as the present time, then use expr to find the difference, and sleep that many seconds.

回答4:

Usually in Linux you use crontab for this kind of scduled tasks. But you have to specify the time when you "setup the timer" - so if you want it to be configurable in the file itself, you will have to create some mechanism to do that.

But in general, you would use for example:

30 1 * * 5 /path/to/script/script.sh

Would execute the script every Friday at 1:30 (AM) Here:

30 is minutes

1 is hour

next 2 *'s are day of month and month (in that order) and 5 is weekday