How to create a simple cron in unix

I need to create a very simple cronjob in unix to be launched at certain time, lets say 22:30, it will need simply to launch a simple script with the command date to be written on a file, just that.


The crons in my system right now are these, in this location:

crontab -l
00 02 * * * $HOME/folder1/script1.ksh >/dev/null
00 15,30,45 * * * * $HOME/folder1/script2.ksh >/dev/null
00 02 * * * $HOME/folder1/script3.ksh > $HOME/menu/log/mylog.log
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celticianAsked:
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woolmilkporcConnect With a Mentor Commented:
30 22 * * * date > /path/to/file

Use "crontab -e" to open the crontab with an editor. Add the above line and save the crontab.

Replace "/path/to/file" with a valid path and name of your choice. Use ">>" instead of ">" to append to the file instead of overwriting it.
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sventhanCommented:
you need to add this entry on your crontab

by typing at the prompt

crontab -e

then add this...

30 22 * * * /path/to/the/script

path to the script is your executable.
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celticianAuthor Commented:
thank you!
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