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I'm using a script for testing broadband speed and I would like to set up a cronjob for testing every n minutes and output to a file. The command to launch it from a shell console and append to logfile prepending a line with the current date is

tespeed.py -w | sed -e "s/^/$(date +\"%d-%m-%y\ %T\"), /" >>tespeedlog.csv

But if I use this command in a cronjob something doesn't work; the syslog reports: 

Sep 25 13:23:01 raspberrypi /USR/SBIN/CRON[6719]: (pi) CMD (/home/pi/tespeed/tespeed.py -w | sed -e "s/^/$(date +')

What shoud I check?

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  • 1
    You need to escape the % characters with backslashes, see your crontab(5) man page for details. Commented Sep 25, 2013 at 12:00
  • Related: unix.stackexchange.com/questions/207/… Commented Sep 25, 2013 at 21:49

1 Answer 1

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Create a wrapper script as there may be a problem with escaping date format. The problem seems to be with % character which may be interpreted as a new line specifier in some cron schedulers:

Put this to the file /usr/local/bin/wrpr.sh:

#!/bin/sh
tespeed.py -w | sed -e "s/^/$(date +\"%d-%m-%y\ %T\"), /" >> /tmp/tespeedlog.csv

Make it executable:

chmod u+x /usr/local/bin/wrpr.sh

And schedule it with cron (this will override the current user's crontab):

echo "* * * * * /usr/local/bin/wrpr.sh" | crontab

Otherwise, use crontab -e to add it to the current user's crontab.

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  • @alzambo if it works, please accept this answer, that's the way to say thanks on the stack exchange sites. Commented Sep 25, 2013 at 14:43

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