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Old 02-25-2011, 06:46 PM   #1
v8625
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Registered: Sep 2002
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need to echo / printf string containing date as parameter in cron


Is it possible to get cron to echo (or printf) the following string:
1110225^1110225^1110225^1110225^1110225
(with 110225 obviously being current date - at command line I obtained this string using
printf 1$(date +"%y%m%d")^1$(date +"%y%m%d")^1$(date +"%y%m%d")^1$(date +"%y%m%d")^1$(date +"%y%m%d")

But cron does not like this string - says
.... printf 1$(date +"
Content-Type: text/plain; charset=ANSI_X3.4-1968
X-Cron-Env: <SHELL=/bin/sh>
X-Cron-Env: <HOME=/home/db2inst1>
X-Cron-Env: <PATH=/usr/bin:/bin>
X-Cron-Env: <LOGNAME=db2inst1>
X-Cron-Env: <USER=db2inst1>

/bin/sh: -c: line 0: unexpected EOF while looking for matching `"'
/bin/sh: -c: line 1: syntax error: unexpected end of file
 
Old 02-25-2011, 07:31 PM   #2
macemoneta
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Registered: Jan 2005
Location: Manalapan, NJ
Distribution: Fedora x86 and x86_64, Debian PPC and ARM, Android
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How about just:

Code:
/bin/date +'1%y%m%d^1%y%m%d^1%y%m%d^1%y%m%d^1%y%m%d'

Last edited by macemoneta; 02-25-2011 at 07:33 PM.
 
  


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