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Old 05-12-2004, 05:47 AM   #1
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Regular expression and tools

I hope I'm posting this message to the correct group.

I need to find the amount of disk space currently occupied in percentages.
This is relatively easy:

/bin/df -k /

It outputs the following:

Filesystem 1K-blocks Used Available Use% Mounted on
/dev/hda1 60476036 9084940 48319068 16% /

From within a bash shell script, I have to get the number 16.
Meaning, I have to look for digits before a percent sign.
I know I have to use regular expressions and something called a backreference but
I haven't been able to accomplish that.

Please provide good links if you know about and a code sample for a solution if possible.
Old 05-12-2004, 06:39 AM   #2
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diskSpace=`/bin/df -k / | perl -ne '/ (\d+)% /; print $1'`
echo "diskSpace=$diskSpace"

if [ $diskSpace -gt 90 ]; then
echo "Disk space is running low."
Old 05-12-2004, 06:55 AM   #3
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Registered: Sep 2003
Location: Sweden
Distribution: Debian
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Without Perl, this is one way of doing the above:
diskspace=`df -k / | grep ^/ | gawk '{ print $5 }' | tr -d '%'`
echo "diskspace = $diskspace"
if [ $diskspace -gt 90 ]
    echo "Disk space is running low."



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