Using variable in awk output

Question

The following works as expected.

awk -F'^' '{printf "set %s:%s %s\n",$1,$2, $7}' todel.txt | sed 's/$/\\r\\n/' >> tofile.txt

But when I try to add a variable to output as shown below, I get an error:

awk -F'^' '{printf "set %s:%s:%s %s\n",$1,$2,$myvar $7}' todel.txt | sed 's/$/\\r\\n/' >> tofile.txt

tripleee · Accepted Answer · 2012-03-15 14:41:21Z

5

$myvar doesn't expand in single quote.
You can use the -v option to pass shell variable to awk:

awk -F'^' -v myvar=$myvar '{printf "set %s:%s:%s %s\n",$1,$2,myvar,$7}' todel.txt

tripleee

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answered Mar 15, 2012 at 14:32

kev

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4 Comments

awk: (FILENAME=todel.txt FNR=1) fatal: not enough arguments to satisfy format string # does not seem to work as expected.

I added the missing comma. I'm not at my computer so cannot test. )-:

Yes, I haven't test the missing comma too.

+1 Best to use quotes for the -v argument to protect whitespace awk -v "myvar=$myvar" ...