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Nawk sub not substituting

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# 8  
Old 06-13-2014
Thanks for the input. I reassigned correctly and tried to alter the command based on the interpretation of the gsub.

I simplified it to the below.
Code:
nawk -F / -vz=Peoccam.20140613 ' { gsub("Peoccam", z,$5) ; print $0} '

However, when I execute the below the return gives the path without the foward slashes -
Code:
smenago@sftp02$ bash -x test_app
++ /usr/local/bin/date +%Y%m%d
+ ydate=20140613
+ echo 20140613
20140613
+ test_ca=/home/mdadmin/test_ca
++ cat /home/mdadmin/test_ca
+ for i in '`cat ${test_ca}`'
+ [[ /pub/data/cana/Peoccam == *cana* ]]
++ /usr/bin/nawk -F/ -v dt=20140613 '{ print $5 "." dt }' /home/mdadmin/test_ca
+ new_f=Peoccam.20140613
+ echo Peoccam.20140613
Peoccam.20140613
+ mv /pub/data/cana/Peoccam /pub/data/cana/Peoccam.20140613
mv: cannot access /pub/data/cana/Peoccam
+ echo /pub/data/cana/Peoccam
/pub/data/cana/Peoccam
+ nawk -F / -vz=Peoccam.20140613 ' { gsub("Peoccam", z,$5) ; print $0} ' /home/mdadmin/test_ca
 pub data cana

# 9  
Old 06-13-2014
There are two mistakes.
1. You must also set OFS to / otherwise the modification of $5, causing a rebuild of $0, uses the default OFS (output field separator is a space).
2. There must be a space between -v and z=

Not important here: gsub tries multiple substitutens, sub is sufficient here.
And there are bugs with brackets if the first argument in sub or gsub is in quotes. Better use slashes as first choice.
Code:
nawk -v z="Peoccam.20140613" 'BEGIN {FS=OFS="/"} {gsub(/Peoccam/,z,$5) ; print}' /home/mdadmin/test_ca

This User Gave Thanks to MadeInGermany For This Post:
smenago
# 10  
Old 06-19-2014
SmilieYes. Thanks for the help as the explanation helped resolve the issue.

Appreciate the help.
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