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allancuntapay 11-06-2009 01:26 AM
home directory size monitoring script
 
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 02:24 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 02:34 AM
Or use quotas, that's kind of what they're for.

indiajoe 11-06-2009 02:37 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 05:51 AM
Thanks a lot. that worked. :)



Quote:
Originally Posted by indiajoe (Post 3746750)
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 06:09 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 06:59 AM
Quote:
Originally Posted by ghostdog74 (Post 3746906)
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.


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