accessing outer shell variables inside nawk in ksh

Question

I am trying to access a shell variable inside a piped nawk. I have never done this before and was wondering if its possible.

Here is the command sbdadm list-lu contents:

Found 2 LU(s)
          GUID                    DATA SIZE           SOURCE
600144f029bf0a0000004e0484740052 107380964864 /dev/rdsk/c9d0s1 600144f029bf0a0000004e0484740053 53694562304 /dev/rdsk/c9d0s3

Here is my sample of my script :

DISK=/dev/rdsk/c9d0s3
sbdadm list-lu |nawk '/$DISK/ {print}'

NOTE: I know the " /$DISK/" syntax will not work since $ is part of a regex symbol. I need the right syntax if such a code is ever possible.

In addition,does awk spawn another shell? If so, is it possible that I can export this variable $DISK to that shell.

Are you aware that the `cat output` is a [UUOC](http://www.catb.org/jargon/html/U/UUOC.html)? — Fred Foo, Jun 24 '11 at 19:37
@larsman: Sorry about the confusion ... basically I wanted sbdadm list-lu> output .... got confused about output being a variable instead of a file ... I have re-corrected the question .. Thanks for the help you guys — tomkaith13, Jun 24 '11 at 19:57

score 1 · Answer 1 · answered Jun 24 '11 at 19:30

export DISK=/dev/rdsk/c9d0s3
cat output | awk '$0 ~ ENVIRON["DISK"]{print}'

results:

600144f029bf0a0000004e0484740053 53694562304 /dev/rdsk/c9d0s3

With system call (after DISK was exported):

echo | awk '{system("echo $DISK")}'

results:

/dev/rdsk/c9d0s3

score 1 · Accepted Answer · answered Jun 24 '11 at 19:43

The problem is not that $ is part of RE syntax; it's that / is the RE delimiter. If you were just looking for c9d0s3, then using the proper quoting would do the trick:

$ DISK=c9d0s3
$ awk "/$DISK/ {print}" output
600144f029bf0a0000004e0484740053 53694562304 /dev/rdsk/c9d0s3

Explanation: if you use "" instead of '', then the shell variable would be expanded before handing the program to awk, so awk would see

/c9d0s3/ {print}

as its program. You can still search for a pattern with / in it, but it takes some shell quoting magic:

$ DISK=/dev/rdsk/c9d0s3
$ awk "{if (match(\$0, \"$DISK\")) print}" output
600144f029bf0a0000004e0484740053 53694562304 /dev/rdsk/c9d0s3

And no, awk does not spawn a subshell. Why would it? And why would you need one to pass a variable if you can just do it through the environment?

Since single quotes dint do the trick for me .. I assumed that the only reason $DISK wasnt visible was coz it was in another shell...Thanks for clarifying — tomkaith13, Jun 24 '11 at 20:03

score 1 · Answer 3 · edited May 23 '17 at 12:13

1

In addition to j.w.r's answer, you can explicitly set an awk variable with the value of the shell variable:

 nawk -v disk="$DISK" '$3 == disk' output_file

edited May 23 '17 at 12:13

Community

1
1

answered Jun 24 '11 at 19:46

glenn jackman

238,783
38
220
352

accessing outer shell variables inside nawk in ksh

3 Answers3