ksh variables inside awk script inside ksh script ?

[Stuart Abramson]

Consider this ksh script:

SIZES= 3 4325 8632 160000

for Size in $SIZES
do
awk '
$NF == $Size { Count = Count + 1
Total = Total + $NF }
END { TotGB = Total / 1000
printf ("%8d %6d %6d GB \n", $NF, Count, TotGB) }
' $SYMDEV.1
done > $SYMDEV.2

I can't get the $Size inside the awk to substitute into a 3, 4315, etc.

How do i do that?

[Mel Burslan]

not an awk/sed expert here but you seem to be missing a "{" in this whole thing. I am not sure if it is the problem, but if not, try escaping the "$" character in front of variable size by a "\", i.e., replace "$Size" with "\$Size"

Hope this helps...

[RAC_1]

awk -v

Anil

[A. Clay Stephenson · Accepted solution]

You are expecting the shell to instantiate variables within single quotes --- that ain't gonna happen but what you could do is something like this:
awk -v Size=${Size} '$NF == Size ....'

The -v Size=${SIZE} create an awk variable 'Size' (not $Size inside awk) and gives it the value of the shell variable ${Size}.

I've intentionally not done everything for you.

[Sandman!]

Stuart,

You need to replace the single ticks with double quotes in your awk script otherwise variable expansion will not occur. Also change the double quotes of your printf statement to single quotes so that awk can correctly parse your statements.

cheers!

[Eknath]

Hi Stuart,

Try this with -v option


SIZES= 3 4325 8632 160000

for Size in $SIZES
do
awk -v NF=$Size '
{ Count = Count + 1
Total = Total + $NF }
END { TotGB = Total / 1000
printf ("%8d %6d %6d GB \n", $NF, Count, TotGB) }
' $SYMDEV.1
done > $SYMDEV.2

Should work

Cheers !!!
eknath

[Stuart Abramson]

awk -v Size=$Size '
$NF == Size {Count = Count +1 ...