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Print if found non-desired result

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# 1  
Old 01-12-2015
Print if found non-desired result
I have a result like this

Code:
root@server [~]# grep -rl maldet /etc/cron*
/etc/cron.d/maldet_daily
/etc/cron.d/malcron
/etc/cron.d/malcrondaily
/etc/cron.d/malcronweekly

What I need is, I need an if/else condition such that, if there is any output other than /etc/cron.d/maldet_daily in the grep command, an "else" condition would say "critical: more than one maldet cron found: /etc/cron.d/malcron -- /etc/cron.d/malcrondaily -- /etc/cron.d/malcronweekly " If only /etc/cron.d/maldet_daily is there in output, it should be "OK:"
# 2  
Old 01-12-2015
Hello anil510,

Could you please try following and let me know if this helps.(Not tested though)
Code:
grep -rl maldet /etc/cron* | awk '{split($0, X, " ");}END{if(length($0)== 1){print "OK"} else {print "CRITICAL"}}'

Thanks,
R. Singh
# 3  
Old 01-12-2015
R singh, Sorry, Its not working
# 4  
Old 01-12-2015
Hello anil510,

Could you please try following and let me know if this helps.
Code:
cat check_cron.ksh
A=`grep -rl  maldet /etc/cron*`
echo $A | awk '{split($0, X, " ");}END{if(length(X)== 1){print "OK"} else if(length(X)==0){print "NO results found."} else {print "CRITICAL"}}'

This should work.

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
anil510
# 5  
Old 01-12-2015
Almost fine it seems. But not the real result.
Code:
root@server [~]#  grep -rl  maldet /etc/cron*
/etc/cron.d/maldetweekly
/etc/cron.d/maldet_daily
/etc/cron.d/malcron
root@server [~]# A=`grep -rl  maldet /etc/cron*` ; echo $A | awk '{split($0, X, " ");}END{if(length(X)== 1){print "OK"} else if(length(X)==0){print "NO results found."} else {print X[1] "CRITICAL"}}'
/etc/cron.d/maldetweeklyCRITICAL

# 6  
Old 01-12-2015
Hello anil510,

Could you please try this and let me know this helps.(Not tested though)
Code:
A=`grep -rl  maldet /etc/cron*`
echo $A | awk '{split($0, X, " ");}END{if(length(X)== 1){print "OK"} else if(length(X)==0){print "NO results found."} else {print "CRITICAL\nResults found are:";for(i in X){print X[i]}}}'

I have mentioned as Results found are: you can remove this if it doesn't require.

Thanks,
R. Singh
# 7  
Old 01-12-2015
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
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