How to automatically generate a file from a script with current date as file name

shellunix

I have some shell script which generates some outputs. The output is written into a file. In my cronjob I have it:
./myscript.sh –options > output
Because the cronjob runs every week, I would like to keep a list of outputs instead of overwritting them every time the script runs. So ideally I would like to have the output file name as output_date. Date is current script running date. In Unix, date command can give you current date info, but how could I integrate that into the output file name?

Update:

Have tried Heath and wfaulk's methods, it does work well. However, when I put the same command in cronjob, I started to get the error:
/bin/sh: -c: line 0: unexpected EOF while looking for matching `)'
/bin/sh: -c: line 1: syntax error: unexpected end of file

I know it's caused by output to a file named output_$(date +%Y%m%d) since that's the only change I made. Is there a different way to do it in cronjob?

Best Answer

myscript.sh > output_$(date +%Y%m%d)

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How to export VARs from a subshell to a parent shell

I think it is a "no can do" sort of problem...

First -- you wouldn't want to source that script because of the exit 0 at the end.

Second, no unix child process can directly change the environment of the parent. Otherwise all sorts of crazy things would be possible.

Could you add something to their environment with the default profile or bashrc files, or could you write a wrapper for whatever program it is they are trying to run?

Allow me to elaborate on the "wrapper" concept.

Let's say you want to run the program snoopy with either PROD or DEV in environment variable "OPTIONS" depending on if you want production or development. IF it isn't set, let's say snoopy does something zany like wipe out the database for production and development...

rename "snoopy" to snoopy.bin (or .snoopy.bin)

then put a script in that same location named "snoopy" that contains this:

#!/bin/sh

export OPTIONS

case "$OPTIONS" 
in
  PROD) ;;
  DEV) ;;
  *) OPTIONS=DEV ;;
esac

#the binary is actually named snoopy.bin 

exec "$0.bin" "$@"

If you don't want to muck with the actual file, put this script somewhere in the filesystem that will be ahead of the actual snoopy program in the user's PATH and have a full path to the binary in the script's exec statement...

Cron – Change the shell for cron job

Why it doesn't Work:
The reason that it doesn't work is that su requires that you run the command from a terminal, and cron does not provide a terminal.

Simple way to do it is using Vixie Cron:
If you didn't mind using Vixie Cron I think the simplest thing to do would be to put the script in for that user by running the following as that user:

crontab -e

Then use 0 0 * * 0 as the time.

If you want to use Anacron as a User:
Anacron is helpful if you want to make sure the job doesn't get missed if the machine was off. To run as different user, specify a different anacron tab file and spool dir, put the following in the system cron to run as the specified user, anacron -t /home/foo/etc/anacrontab -S /home/foo/var/spool/ and then edit that tab file to run the script.

There is also a mini-howto, but this looks more complicated.

You will have to test these though to verify. Lastly, remember this for future testing run-parts --test /etc/cron.weekly

Best Answer

Related Solutions

How to export VARs from a subshell to a parent shell

Cron – Change the shell for cron job

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