AWK打印$ 2 Breaking Bash Args

Question

I'm trying to capture the output from an EC2 command with AWK. The AWK portion works and the EC2 command work. The problem is, I pass arguments to the script and one of them conflicts with the style of AWK. Specifically print $2

Proper segment is

cmd="/opt/aws/bin/ec2-run-instances -O $secid  -W $seckey $ami  -n $1 -g $secg -k $sshkey -t $instsize -z $2 | awk '/^INSTANCE/ {print $2}'

As you can see, I need that print $2 to capture the EC2 instance id. Is there a workaround without changing my arg format?

Thanks!

RESERVATION     r-******      ********    www.abc.com
INSTANCE        i-****      ami-*****

Answer 1

I believe you just need to escape it for awk so that shell doesn't replace $1 , $2 etc, liek this:

awk '/^INSTANCE/ {print \$2}'

OR your CMD variable:

cmd="/opt/aws/bin/ec2-run-instances -O $secid  -W $seckey $ami  -n $1 -g $secg -k $sshkey -t $instsize -z $2 | awk '/^INSTANCE/ {print \$2}'"

Answer 2

result=$(/opt/aws/bin/ec2-run-instances -O $secid  -W $seckey $ami  -n $1 -g $secg -k $sshkey -t $instsize -z $2 | awk '/^INSTANCE/ {print $2}')
for instance_id in $result; do echo $instance_id; done

Answer 3

尝试这样的事情：

cmd="/opt/aws/bin/ec2-run-instances -O $secid  -W $seckey $ami  -n $1 -g $secg -k $sshkey -t $instsize -z $2 | awk -v var=$2 '/^INSTANCE/ {print var}'