DF and AWK Command

ntbluez · 02-02-2016, 09:52 AM

I want to run a command (make it a script) to show me which volumes that are 20% or higher in usage. I am having problem with following syntax as it is not showing me which volume is 20%+ usage.

Appreciate your help!

df | awk '$1 $5~/([2-9]|10)[0-9]/ {exit 1}' && echo 'No >=20s' || echo 'Over 20% Usage!'

rknichols · 02-02-2016, 10:59 AM

Without seeing the results from df this is just a guess, but you might need to use "df -P" to ensure that long device names don't cause output to be split into two lines.

ntbluez · 02-02-2016, 11:55 AM

resulting output is:
Over 20 Usage!

It does not list the volume that is 20% or higher & that is what I want listed in result/output.

rknichols · 02-02-2016, 01:49 PM

What is the output when you run "df" alone, and compare that to the output from "df -P".

ondoho · 02-02-2016, 03:04 PM

what is the purpose of the '{exit 1}'?
the way i see it, awk will always exit with 1 (=false), and therefore the combined A && B || C structure will always choose C. which results in output of "Over 20 usage".

ntbluez · 02-02-2016, 03:29 PM

If it does, it exits code 1 and the || code executes (there is a drive over 20 used) and If it doesn't, it exits code 0 and the && code executes. I'm looking to list the volume which is 20 or more utilized.

In this example, I want DF output to show: /dev/sda1 80% and /dev/sda2 50%.

A typical DF output is

Filesystem 1K-blocks Used Available Use% Mounted_On
/tmpfs 370692 0 370692 0% /dev
/dev/sda1 19601408 15681126 3920281 80% /
/dev/sda2 1000 500 500 50% /boot

rknichols · 02-02-2016, 06:08 PM

OK, so "-P" isn't the issue at all, though you really should use it so that output like this doesn't cause confusion:

Code:

Filesystem           1K-blocks      Used Available Use% Mounted on
/dev/mapper/vg_omega3g-rootvol
                      11963960   5975292   5357884  53% /
tmpfs                  8194160       204   8193956   1% /dev/shm

The line break, which would not occur with "-P", makes that harder to parse.

Exactly what in your original code do you think is going to print the volume name?

Quote:

Code:

df | awk '$1 $5~/([2-9]|10)[0-9]/ {exit 1}' && echo 'No >=20s' || echo 'Over 20% Usage!'

The awk command doesn't print anything at all, and the echo commands just print the text you have there. What do you want to see if there is more than one volume with >20% usage, or are you perhaps not concerned about that? Right now, the awk script exits as soon as it sees the first one, so it would have to be changed to make it see more than one. Are you looking for output all on one line, e.g.:

Code:

volume1 [volume2 ...] Over 20% usage!

or would one or more lines with just volume names suffice? Easiest is a separate line with just the first oversized volume name:

Code:

df | awk '$1 $5~/([2-9]|10)[0-9]/ {print $1; exit 1}' && echo 'No >=20s' || echo 'Over 20% Usage!'

syg00 · 02-02-2016, 06:20 PM

exit stops the run - therefore no print of the record. And even if it did, you'd only ever get one record, not all over 20.

There is rarely any need to shell out of awk to process its output by other tools. awk is more than capable of handling all you require in this case. As for that regex, that is only needed in cases where the tool doesn't have arithmetic comparison operators. awk does have those operators, so use that instead.

ntbluez · 02-02-2016, 09:20 PM

You Are Awesome!!!

One more, what if I wanted to also include the hostname? Meaning, I am using a listing of server names to run this script against, and I was wondering how to include that hostname while using this DF command in a SSH session...?

ssh MyUserID@ServerName df | awk '$1 $5~/([2-9]|10)[0-9]/ {print $1; exit 1}' && echo 'No >=20s' || echo 'Over 20 Usage!' && hostname

I was thinking something like adding additional command like "&& hostname" but having no luck as it produces my local host name.

Thanks again!!!

ondoho · 02-03-2016, 02:05 AM

Quote:

Originally Posted by ntbluez

I was thinking something like adding additional command like "&& hostname" but having no luck as it produces my local host name.

so what hostname do you want then? not sure i understand.

PS:
if you're actually using this as a script, meaning it is saved as an executable file, i would rather not use the oneliner-style with && and ||, but write it properly multiline instead.
execution time will be the same (if you do it right of course), but you have much more control and options.

ntbluez · 02-03-2016, 06:52 AM

@Ondoho

Yes. I want to run these commands via script. If you suggest I do so, what would the script look like? I know I have a file called SERVERS which lists those servers that ssh uses to make each connection. I want the "Hostname" of that server to be included in output of that DF command...

So, lets say I have a file called SERVERS with three (3) server names; server1.com server2.com and server3.com.
Requirement: need SSH make connection to each server and perform these commands to collect information about that server (a) list any disk volume greater than or equal to 70 utilization (b) list % of usage for that volume 70%+ (c) provide that server hostname. I want results to be in an output file on my local PC or server which I run script from.

Space.sh...
#!/bin/sh
for SERVERS in 'cat server-list'
do
ssh MyID@$SERVERS df | awk '$1 $5~/([2-9]|10)[0-9]/ {print $1; exit 1}' && echo 'No >=20s' || echo 'Over 20% Usage!' && grep $HOSTNAME=* >> /home/joe/script/something.txt
done

rknichols · 02-03-2016, 08:40 AM

Just pass the server name to awk as a variable and include that variable in the print statement:

Code:

awk -v "Server=$SERVER" '$1 $5~/([2-9]|10)[0-9]/ {print Server, $1; exit 1}'

ondoho · 02-03-2016, 03:36 PM

i think there's a problem with your script; at least on my system it spits out only the first partition that is over 20% usage; although most of my partitions are.

anyhow, i felt like transforming this into a proper script:

Code:

#!/bin/bash

while read line
do
perc="${line%%\%*}"
if (( perc > 19 ))
then
echo $line
fi
done <<<"$(df --output=pcent,source)"

to understand, read

Code:

man df

and this:

Quote:

Originally Posted by http://www.tldp.org/LDP/abs/html/string-manipulation.html

${string%%substring}

Deletes longest match of $substring from back of $string.

stringZ=abcABC123ABCabc
echo ${stringZ%%A*}
$ abc