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Print if found non-desired result

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# 8  
Old 01-12-2015
Quote:
Originally Posted by anil510
I have found the solution.

Code:
multimalcron=`grep -rl  maldet /etc/cron* | grep -v maldet_daily`
 
if [ "$(echo $multimalcron)" ]; then
                echo " Multiple maldet cron found: echo $multimalcron"
fi

Hello anil510,

As per your requirement in Post#1, you have asked that you need to print ok when it is only single entry in cron, the code which is provided by you will not satisfy that condition, also I have added a condition if no entry found for maldetthen it will show that. Could you please try code which I provided you in Post#6, let me know if that helps you.


Thanks,
R. Singh
Last edited by RavinderSingh13; 01-12-2015 at 03:54 AM.. Reason: Added 2 icode tags
# 9  
Old 01-12-2015
The following seems to provide something closer to the spirit of what was requested in the original post in this thread (and, except for the one call to grep, only uses shell built-ins):
Code:
list="$(grep -rl 'maldet' /etc/cron*)"
if [ "$list" = "/etc/cron.d/maldet_daily" ]
then	echo 'OK:'
else	if [ "$list" = '' ]
	then	echo 'critical: maldet not found:'
	else	printf 'critical: maldet cron found in the following file(s): '
		last=''
		echo "$list" | while read line
		do	if [ "$last" != '' ]
			then	printf '%s -- ' "$last"
			fi
			last="$line"
		done
		printf '%s\n' "$last"
	fi
fi
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