[SOLVED] Using script variable in awk subtraction

theillien · 05-29-2012, 05:58 PM

I have the following script:

Code:

#!/bin/bash

##################################################################
#
# Purpose:
# When RHEL6 VMs are moved between compute pools or VLANs their
# MAC addresses change breaking the interfaces. This script 
# updates the ifcfg-eth* and network interface udev rules files
# to make the configuration match the new addresses.
#
# Changelog:
# 1.0 - Initial script - 05/26/2012 - msnyder
# 1.1 - Switched $udevpath to $udevfile incorporating the actual
#       file in the variable since it is always the same; added
#       documentation - 05/27/2012 - msnyder
# 2.0 - Rewrite to account for variable number of improper
#       udev file entries - 05/29/2012 - msnyder
#
##################################################################


#Set paths to file locations
udevfile=/etc/udev/rules.d/70-persistent-net.rules
netpath=/etc/sysconfig/network-scripts
ifaces=$(grep eth $udevfile |wc -l)
	
shft=$(($ifaces - 1))
adj=$(($ifaces - 3))

for (($i=0; $i<=$shft; $i++))
do
        #Extract MAC addresses of old and new udev entries
        #These will be used to update the ifcfg-eth* files
        MAC[$i]=$(grep eth$i $udevfile |awk -F ',' '{print $4}' |awk -F "==" '{print $2}')

	if [ $i -lt 3 ]
	then
		#If $i is less than 3 comment out the eth$i entry in the udev rules file
		sed -i "/eth$i/ s/^/#/" $udevfile
	elif [ $i -ge 3 ] && [ $i -lt $adj ]
	then
		#Delete the lines associated with the old "new" interfaces
		sed -i "/eth$i/d"
	else
		#Adjust the eth field to the correct value
		int=$(echo $i |awk -v val=$adj '{print ($1 - $val)}')
		sed -i "s/eth$i/eth$int/" $udevfile

		#Change the MAC address in the ifcfg-eth* files to match
		#the entry in the udev rules file for the new interfaces
		sed -i "s/HWADDR.*/HWADDR=${MAC[$i]}/" $netpath/ifcfg-eth$int
	fi
done

reboot

The issue is specifically with the line that assigns the output of the awk subtraction to the int variable:

Code:

int=$(echo $i |awk -v val=$adj '{print ($1 - $val)}')

I know that $val is being assigned the value of $adj as is evidenced by the screenshot.* However, the subtraction does not execute and the value of $int is not set properly (for an $adj of 5 I expect an $int of 2).

What am I doing wrong with this line that it isn't subtracting the value properly?

*I had to use a screenshot because the VMware console window doesn't support copying/pasting text.

theillien · 05-29-2012, 06:22 PM

I think I solved this. Instead of using the '-v val=$adj' assignment I used

Code:

int=$(echo $i |awk '{print ($1 - '$adj')}')

I'm now seeing the results I expect.

grail · 05-30-2012, 03:42 AM

Firstly, your issue was the you used $adj in awk where you call variables with their name, ie. just 'adj'.

Secondly, you previously use bash math:

Code:

adj=$(($ifaces - 3))

Is there a reason to not use it here?

Also you can simplify your current usage further:

Code:

(( adj = ifaces - 3 ))

pan64 · 05-30-2012, 03:55 AM

the problem with that awk is that you do not need $ inside the script:
int=$(echo $i |awk -v val=$adj '{print ($1 - $val)}')
you need to remove it - but of course, (( int = i - adj )) is much better

in line: MAC[$i]=$(grep eth$i $udevfile |awk -F ',' '{print $4}' |awk -F "==" '{print $2}')
only one awk would be enough instead of grep|awk|awk

theillien · 05-30-2012, 08:24 AM

What would the one awk in the MAC line look like?

pan64 · 05-30-2012, 08:31 AM

awk -F, " /eth$i/ "' { split ($4, a, "=="); print a[2] }'
I have not tested because I have no input file, but probably works, after testing:
MAC[$i]=$( this is the awk )
if you give a sample file we can try to test it also

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grail · 05-30-2012, 09:15 AM

I would also point out that your for loop is not going to work either as you are trying to assign the value 0 to the value of 'i' by using $i.

Also, you asked previously how to get the MAC values in your question here, which showed you a number of better methods.

Code:

ifaces=$(grep eth $udevfile |wc -l)

wc is an over kill here, try looking up the -c option for grep.

As with the math examples above you could also change your tests from the older [] form to also use (()):

Code:

if [ $i -lt 3 ]

if (( i < 3 ))

theillien · 06-11-2012, 09:25 AM

Thanks for the additional input. I discovered the assignment of 0 to $i when I started putting this into full test mode.

Thanks for the heads up with the -c option for grep. That should provide some more elegant lines in the future.