[SOLVED] home directory size monitoring script

allancuntapay · 11-06-2009, 02:26 AM

newbie here. am doing script on monitoring my directory /home that if it gets greater or equal to 86, then it will prompt me overload, else its fine.

please help what's wrong with my script. thanks

SIZE='df -h /home | /bin/awk '{ print $5 }' | grep -v Use% | sed '/%//''
if [ $SIZE >= "86" ]; then
echo "overload"
else
echo "ok"
fi

Allan

jlinkels · 11-06-2009, 03:24 AM

One obvious error is $SIZE > "86" which should be $SIZE -gt 86.

Another problem might be grep Use%, I would grep Use\%. In Linux everything appearing on top of the number keys is suspicious and might need to be escaped.

Finally run you script with sh -x yourscript

jlinkels

tuxdev · 11-06-2009, 03:34 AM

Or use quotas, that's kind of what they're for.

indiajoe · 11-06-2009, 03:37 AM

Hi,
Following is the corrected code. Another error was that you used /%// instead of s/%// in sed .

Code:

#!/bin/sh
SIZE=`df -h /home/YOU | /bin/awk '{ print $5 }' | grep -v Use% | sed 's/%//'`
if [ $SIZE -ge 86 ]; then
echo "overload"
else
echo "ok"
fi

-Cheers
indiajoe

allancuntapay · 11-06-2009, 06:51 AM

Thanks a lot. that worked.

Quote:

Originally Posted by indiajoe

Hi,
Following is the corrected code. Another error was that you used /%// instead of s/%// in sed .

Code:

#!/bin/sh
SIZE=`df -h /home/YOU | /bin/awk '{ print $5 }' | grep -v Use% | sed 's/%//'`
if [ $SIZE -ge 86 ]; then
echo "overload"
else
echo "ok"
fi

-Cheers
indiajoe

ghostdog74 · 11-06-2009, 07:09 AM

do it with just awk. No need grep or sed

Code:

$ df /home | awk '$5+0>86{print "overload"}'
overload

allancuntapay · 11-07-2009, 07:59 AM

Quote:

Originally Posted by ghostdog74

do it with just awk. No need grep or sed

Code:

$ df /home | awk '$5+0>86{print "overload"}'
overload

thank you very much.. Am learning a lot in here.