Print if found non-desired result

Shell Programming and Scripting
1

I have a result like this

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

Hello anil510,

Could you please try following and let me know if this helps.(Not tested though)

grep -rl maldet /etc/cron* | awk '{split($0, X, " ");}END{if(length($0)== 1){print "OK"} else {print "CRITICAL"}}'

Thanks,
R. Singh

3

R singh, Sorry, Its not working

4

Hello anil510,

Could you please try following and let me know if this helps.

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

1 Like
5

Almost fine it seems. But not the real result.

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

Hello anil510,

Could you please try this and let me know this helps.(Not tested though)

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

Thanks,
R. Singh

7

I have found the solution.

multimalcron=`grep -rl  maldet /etc/cron* | grep -v maldet_daily`

if [ "$(echo $multimalcron)" ]; then
                echo " Multiple maldet cron found: echo $multimalcron"
fi
8

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 maldet then 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

9

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):

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