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print user variable input

 
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# 1  
Old 10-04-2012
print user variable input
this may be basic thing for everyone here, but i cant push awk to print the variable user input which is INS, please help.

code:
Code:
INS=$1

printf '\n'
symdg list | grep $INS-clone | awk -v i=$INS '{ID=substr($4,9,4)}{print "Device Group: "$1,"at Array "ID,i}'

output:
Code:
$ ./test.sh patty

awk: syntax error near line 1
awk: bailing out near line 1

# 2  
Old 10-04-2012
try using nawk
# 3  
Old 10-04-2012
it worked, thanks! what would be difference / specific application wherein nawk is better than awk??
# 4  
Old 10-04-2012
Quote:
Originally Posted by prodigy06
what would be difference / specific application wherein nawk is better than awk??
Are you using Solaris..?

As per my findings /usr/pkg4/bin/awk and nawk used in Solaris..

please do let me correct if i am wrong. not familiar with Solaris..Smilie
Last edited by pamu; 10-04-2012 at 06:10 AM..
# 5  
Old 10-04-2012
yes, its Solaris 9
# 6  
Old 10-04-2012
You could also put variables into awk the old-fashioned way:

Code:
awk '{...}' VARNAME="$VAR"

You really should be double-quoting all your shell variables, by the way.

Also, if you're using awk, you really don't need grep.

Code:
symdg list | awk '$0 ~ I"-clone" { ID=substr($4,9,4); print "Device Group: "$1,"at Array "ID,I}' I="$INS"

Last edited by Corona688; 10-04-2012 at 01:27 PM..
 
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