I need to substitute a portion of an awk command in a script with different cases, from this:


RESULTS=$(ssh -o LogLevel=QUIET -t -i ~/key.pem user@server sudo docker stats --no-stream $CONTAINER | awk '{if (NR!=1) {print substr($2, 1, length($2)-1)}}')

to this:



case "${TYPE}" in   
        AWK="'{if (NR!=1) {print substr($2, 1, length($2)-1)}}'"

RESULTS=$(ssh -o LogLevel=QUIET -t -i ~/key.pem user@server sudo docker stats --no-stream $CONTAINER | awk $AWK)

But I always get a syntax error.

  • Does initial expression work good? Aug 23, 2016 at 10:01
  • You need to escape the $2 since it will be interpreted by the shell before awk.
    – user14755
    Aug 23, 2016 at 10:02

1 Answer 1


In a variable declaration, anything (other than parameter/command/arithmetic expansion) inside " is treated literally, so when you do:


$var will be expanded to 'foobar', not foobar.

So you need:

AWK='{if (NR!=1) {print substr($2, 1, length($2)-1)}}'

as you don't need any pre-expansion by shell, better to use single quotes to stay on the safer side. If you use double quotes, then e.g. $2 will be expanded beforehand by the shell as parameter expansion, this is true for any other tokens that are special to shell.

And don't forget to quote $AWK when you use it to disable the split+glob operator:

… | awk "$AWK"
  • OK thanks, my problem was the "" to $AWK, thanks!
    – Steph
    Aug 23, 2016 at 10:27

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