[SOLVED] run from cron (echo to file) 2.6.32-400.21.1.el5uek x86_64 GNU/Linux

oraenthu@live.com · 03-03-2014, 02:58 PM

Hi,
the following file which resides in /home/oracle/cronscripts/batchjobs/ is being run from cron

Code:

SCRIPTS_DIR=$DIR/scripts
SCR_BASE=$HOME/cronscripts
SCR_LOC=$SCR_BASE/batchjobs
LOG_LOC=/$HOME/cronscripts/batchjobs/
LOG_FILE=$LOG_LOC/test_log.log
PATH=/usr/local/bin:/bin:/usr/bin
export PATH
echo ' **************************************************************************************************** ' >$LOG_FILE
echo ' **************************************************************************************************** ' >>${LOG_FILE}
echo " stoping process...${0} " >> ${LOG_FILE}
echo ' Time : ' `date` >> ${LOG_FILE}
echo " Executing  ...${0} " >> ${LOGFILE}
echo " Hello World ..........." >> ${LOG_FILE}
starttime=`date`
echo ' **************************************************************************************************** ' >>$LOG_FILE

This is what I get

Code:

 **************************************************************************************************** 
 **************************************************************************************************** 
 stoping process.../home/oracle/cronscripts/batchjobs/tests.sh 
 Time :  Mon Mar 3 14:32:01 CST 2014
 Hello World ...........
 ****************************************************************************************************

crontab -l|tail -3

Code:

32 14 * * * /bin/sh /home/oracle/cronscripts/batchjobs/tests.sh

the line
echo " Executing ...${0} " >> ${LOGFILE}
does not ever execute
Please help me with some clues/suggestions/answers
Any help is appreciated

colucix · 03-03-2014, 03:00 PM

Code:

echo " Executing  ...${0} " >> ${LOGFILE}

You missed the underscore in the variable name: it should read LOG_FILE instead of LOGFILE.

oraenthu@live.com · 03-03-2014, 03:05 PM

Oh my God! This almost sucked the life out of me.
Thanks so much..
I am crying out of happiness
:-)